r-NRLF 


EOMETRY 


LENNES 


IN  MEMORIAM 
FLOR1AN  CAJOR1 


SOLID  GEOMETRY 


WITH 


PROBLEMS  AND  APPLICATIONS 


BY 


H.    E.    SLAUGHT,    PH.D. 
It 

ASSOCIATE   PROFESSOR   OF    MATHEMATICS   IN   THE   UNIVERSITY 
OF   CHICAGO 

AND 

N.   J.    LENNES,    PH.D. 

INSTRUCTOR   IN    MATHEMATICS    IN   COLUMBIA   UNIVERSITY 


Boston 

ALLYN   AND   BACON 
1911 


COPYRIGHT,    1911, 
BY  H.    E.   SLAUGHT 
AND   N.  J.  LENNES. 


J.  8.  Cushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


PREFACE. 

THE  important  features  by  which  the  Solid  Geometry  seeks 
to  accomplish  the  two  main  purposes  stated  in  the  preface 
of  the  Plane  Geometry  are : 

1.  The  concepts  of  three-dimensional  space   are  made  clear 
by  many  simple  illustrations  and  exercises.     At  best  the  average 
pupil  comes  but  slowly  and  gradually  to  a  full  comprehension 
of  space  forms  and  relations.     Pages  4,  5,  8,  10,  11,  17,  18, 
etc.,  will  show  how  the  pupil  is  aided  in  understanding  and 
appreciating  these  forms  and  relations  at  the  outset.     In  this 
connection,  Chapter  VI  on  Graphic  Representation  is  of  funda- 
mental importance,  since  it  exhibits  to  the  eye  the  functional 
relations  among  varying  geometric  forms. 

2.  Interesting  concrete  applications  are  interspersed  throughout 
the  text.     It  is  the  aim  profoundly  to  impress  upon  the  mind 
of  the  pupil  the  practical  significance  of  certain  fundamental 
theorems  in  solid  geometry.     For  instance,  how  many  pupils  in 
the  ordinary  study  of  the  theorems  on  the  ratios  of  surfaces 
and  volumes  of  similar  figures  realize  that  there  is  in  them 
any  connection  with  the  possibility  of  successfully  launching 
a  steamship  a  mile  long  ?     See  the  exercises  on  pages  112,  113, 
166,  167.      For  illustrations  of  other  interesting  and  useful 
applications,  see  pages  59,  66,  67,  156,  157. 

3.  The   logical   structure   is  made  more   complete   and  more 
prominent  than   in   the   Plane  Geometry.      Solid  Geometry  is 
studied  by  more  mature  pupils  who  have  been  led  by  grad- 
ual stages  in  the  Plane  Geometry  to  a  knowledge  and  appre- 

iii 


OO 


iv  PREFACE. 

elation  of  deductive  reasoning.  Hence  the  axioms  are  stated 
and  applied  in  strictly  scientific  form  and  at  the  precise  points 
where  they  are  to  be  used.  For  instance,  see  §§  5,  7,  9,  83, 
105,  107,  138,  157,  etc. 

Note  also  the  consistent  and  scientific  definitions  of  all 
solids,  not  as  bounded  portions  of  space,  but  as  configurations 
in  space,  the  uniform  conception  in  all  higher  mathematical 
usage.  See  §§  68,  70,  99,  132,  149,  200. 

4.  Throughout  both  parts  of  this  Geometry  a  consistent  scheme 
has  been  followed  in  the  presentation  of  incommensurables  and 
the  theory  of  limits.  In  Chapters  I  to  VI  of  the  Plane  Geome- 
try the  idea  of  "  approach "  is  made  clear  by  many  concrete 
illustrations,  and  the  theorems  involving  this  idea  are  shown 
to  hold  for  all  possible  approximations.  In  the  final  chapter 
of  the  Plane  Geometry  rigorous  proofs  of  these  theorems  are 
given  and  in  far  simpler  terminology  than  is  found  in  current 
text-books.  In  the  Solid  Geometry  this  latter  method  is 
followed  throughout  Chapters  I  to  VI,  thus  giving  a  complete 
and  scientific  treatment  up  to  that  point.  In  Chapter  VII 
of  the  Solid  Geometry  the  theory  of  limits  is  presented  in 
such  a  way  as  to  leave  nothing  to  be  unlearned  or  compromised 
in  later  mathematical  work.  This  chapter  may  be  omitted 
without  affecting  the  logical  completeness  of  the  book. 

H.    E.    SLAUGHT. 
N.  J.   LENNES. 
CHICAGO  AND  NEW  YORK, 
March,  1911. 


CONTENTS. 

CHAPTER  I.     Lines  and  Planes  in  Space. 

PAGE 

Determinations  of  a  Plane          ........  1 

Theorems  on  Perpendiculars      ........  6 

Theorems  on  Parallels        .         .                 ...         ...  12 

Dihedral  Angles 18 

Polyhedral  Angles 26 

Summary  of  Chapter  I .28 

Problems  and  Applications         ........  28 

CHAPTER  II.     Prisms  and  Cylinders. 

Definitions ,         .32 

Theorems  on  Prisms  ..........  34 

Volumes  of  Rectangular  Parallelepipeds 39 

Volumes  of  Prisms  in  General    .......        ..  42 

Cylinders.     Definitions      .        .         .         ....  46 

Measurement  of  the  Surface  and  Volume  of  a  Cylinder     ...  50 

Problems  and  Applications 53 

Theorems  on  Projection 55 

Projection  of  a  Plane  Segment 62 

Summary  of  Chapter  II      .........  65 

Problems  and  Applications         ........  66 

CHAPTER  III.     Pyramids  and  Cones. 

Definitions  and  Theorems  on  Pyramids     ......  68 

Definitions  and  Theorems  on  Cones 80 

Measurement  of  the  Surface  and  Volume  of  a  Cone           ...  84 

Summary  of  Chapter  III 90 

Problems  and  Applications         .        .         .        .         .         .         .         .91 

CHAPTER  IV.     Regular  and  Similar  .Polyhedrons. 

Regular  Polyhedrons 95 

Inscription  of  Regular  Polyhedrons    .         .         .         .         .  99 


VI  CONTENTS. 

PAGE 

Similar  Polyhedrons 102 

Applications  of  Similarity 110 

Summary  of  Chapter  IV 112 

Problems  and  Applications 112 

CHAPTER  V.     The  Sphere. 

Plane  Sections  of  the  Sphere 114 

Trihedral  Angles  and  Spherical  Triangles          .....  124 

Polar  Triangles 132 

Polyhedral  Angles  and  Spherical  Polygons         .....  140 

Areas  of  Spherical  Polygons 141 

Area  and  Volume  of  the  Sphere 145 

Summary  of  Chapter  V      .........  155 

Problems  and  Applications         ........  156 

CHAPTER  VI.     Variable  Geometric  Magnitudes. 

Graphic  Representation 158 

Summary  of  Chapter  VI 165 

Problems  and  Applications         ...         .....  166 

CHAPTER  VII.     Theory  of  Limits. 

General  Principles      .         .         ........  168 

Application  of  Limits  to  Geometry 172 

Volume  of  the  Sphere         .........  182 

Summary  of  Chapter  VII  .........  185 

INDEX 187 


SOLID   GEOMETRY. 

CHAPTER   I. 

LINES  AND  PLANES  IN  SPACE. 

1,  A  figure  in  plane  geometry  is  restricted  so  that  all 
its  points  lie  in  the  same  plane.     Such  figures  are  called 
two-dimensional  figures. 

A  straight  line  is  a  one-dimensional  figure,  and  a  point  has  no 
dimensions. 

2,  A  three-dimensional  figure  is  a  combination  of  points, 
lines,  and  surfaces  not  all  parts  of  which  lie  in  the  same 
plane. 

E.g.  the  six  surfaces  of  the  walls,  floor,  and  ceiling  of  the  school- 
room form  a  three-dimensional  figure.  This  figure  is  not  the  room  itself. 
The  room  is  the  space  inclosed  by  the  figure. 

3,  Solid   geometry  treats   of   the   properties   of    three- 
dimensional  figures. 

DETERMINATION  OP  A  PLANE. 

4,  Since  we  are  to  consider  points  and   lines  not  all 
lying  in  the  same  plane,  it  is  of  first  importance  to  be  able 
to  distinguish  one  plane  from  another. 

This  is  all  the  more  important,  since  three-dimensional  figures  have 
to  be  represented  by  pencil  or  crayon  drawings  on  the  plane  of  the 
paper  or  blackboard.  Models  of  paper,  wire,  etc.,  may  be  constructed 
by  the  pupils. 


SOLID   GEOMETRY. 

5,  Axiom  I.     If  two  points  of  a  straight  line  lie  in 
a  plane,  the  whole  line  lies  in  the  plane. 

Since  a  line  is  endless,  it  follows  from  this  axiom  that  a  plane  is 
unlimited  in  all  its  directions. 

6,  While  two  points  determine  a  line,  this  is  not  suffi- 
cient in  the  case  of  a  plane. 

E.g.  suppose  a  plane  L  contains  the  points 
A  and  B  which  determine  the  line  AB.  If 
the  plane  L  be  revolved  about  the  line  A  B 
as  an  axis,  it  may  occupy  indefinitely  many 
positions,  as  L,  M,  N,  but  there  is  only  one 
position  in  which  it  contains  a  third  given 
point  C  outside  of  the  line  AB. 

7,  Axiom  II.     Through  three  points  not  all  in  the 

same  straight  line  one  and  only  one  plane  can  be  passed. 

8,  THEOKEM.     A  plane  is  determined  by  (1)  a  line 
and  a  point  not  in  that  line,  (2)  two  intersecting  lines, 
(3)  two  parallel  lines. 


i         B^    I  IM       g 

^J^    /,/  /       _ 

^A     G     C~  I  I        A           B 

""  " 


Proof  :   (1)  Let  /  be  the  given  line  and  P  the  point  outside. 

Take  any  two  points  A  and  B  on  L  Then  A,  .B,  and  P 
determine  a  plane,  which  we  call  M  (Ax.  II). 

Take  any  other  two  points  C  and  D  on  1.  Then  C,  D, 
and  P  determine  a  plane,  which  we  call  N  (Ax.  II). 

By  Ax.  I,  N  contains  the  points  A  and  B.  Hence  N  is 
the  plane  determined  by  A,  B,  and  P,  since  it  contains 
these  points.  That  is,  N  is  the  same  plane  as  M. 


2>f 

LINES  AND  PLANES  IN  SPACE.  3 

Therefore  one  and  only  one  plane  is  determined  by  a 
line  and  a  point  not  on  that  line. 

(2)  Let  /!  and  72  be  two  intersecting  lines. 

Take  A  the  intersection  of  ^  and  Z2,  and  B  and  C  any 
other  points,  one  on  Zx  and  the  other  on  12. 

Then  4,  B,  and  C  determine  a  plane  M  in  which  both  ?x  and 
12  lie,  since  J.  and  B  lie  on  ^  and  A  and  C  on  12  (Ax.  I). 

Now  Jf  is  the  only  plane  in  which  both  ^  and  ?2  lie,  for 
any  other  three  points  E,  F,  G,  on  Z:  and  12  (not  all  in  the 
same  line)  determine  the  same  plane  Jf,  since  they  all  lie 
in  it. 

Hence,  two  intersecting  lines  determine  a  plane. 

(3)  Let  /!  and  72  be  two  parallel  lines. 

By  definition  Zx  and  12  lie  in  a  plane  M. 

Now  M  is  the  only  plane  in  which  both  l±  and  Z2  lie,  for 
any  three  points  A,  .B,  C,  on  Z:  and  Z2  (not  all  on  the  same 
line)  determine  this  plane  Jf,  since  they  all  lie  in  it. 

Hence,  two  parallel  lines  determine  a  plane. 

9,   Axiom  III.     If  two  planes  have  a  point  in  com- 
mon, then  they  have  at  least  another  point  in  common. 

10,  THEOREM.  Two  intersecting  planes  have  a 
straight  line  in  common,  and  no  points  in  common 
outside  of  this  line. 

Proof  :  If  two  planes  intersect,  they  must  have  at  least 
two  points  in  common  (Ax.  III). 

But  these  two  points  determine  a  straight  line  which 
lies  wholly  in  each  plane  (Ax.  I). 

Hence,  there  is  a  straight  line  common  to  the  two  planes. 

Now  prove  that  these  two  planes  can  have  no  point  in 
common  outside  of  this  line  unless  the  planes  are  identical. 


4 


SOLID   GEOMETRY. 


11. 


EXERCISES. 


The  diagram  on  this  page  represents  a  three-dimensional  figure  in 
the  shape  of  an  ordinary  box.  In  this  figure  the  points  A,  K,  B  do 
not  determine  a  plane,  since  they  all  lie  on  the  same  straight  line, 
while  A,  C,  E  do  not  lie  in  a  straight  line,  and  hence  determine  a 
plane. 

1.  In  this  figure  pick  out  twelve  sets  of  three  points  each  which 
do  not  determine  planes,  and  also  twelve  sets  which  do  determine 
planes. 

H. G 


2.  Describe  the  relation  to  the  three-dimensional  figure  of  each 
plane  determined  in  Ex.  1.     Thus,  the  plane  WXY  cuts  off  the  lower 
right-hand  corner  at  the  back.  . 

3.  Read  each  of  the  six  bounding  planes  in  this  figure,  using  vari- 
ous different  sets  of  points  or  lines.     Thus,  the  face  ABFE  is  deter- 
mined by  the  points  B,  L,  K,  or  by  A  and  the  line  PN,  or  by  lines  AB 
and  EF. 


LINES  AND  PLANES  IN  SPACE. 


4.  Describe  the  position  of  each  of  the  following  planes  in  the 
figure:  AEC,  ACF,  BCE,  HCA,  ADG,  BDF,  HEC. 

5.  Pick  out  six  planes,  each  determined  by  two  parallel  lines  in 
the  figure. 

6.  Decide  whether  each  of  the  six  planes  in  Ex.  5  may  also  be 
determined  by  two  intersecting  lines  of  the  figure ;  by  a  line  and  a 
point  outside  of  it. 

7.  Using  the  point  Z,  pick  out  six  planes,  each  determined  by 
it  and  two  other  points  of  the  figure,  as  AZD,  AZB,  etc. 

8.  Using  the  schoolroom  or  a  room  at  home,  imagine  all  the  planes 
constructed   which  are  involved  in  the  forego- 
ing questions.     Do  the  same,  using  a  small  box 

or  paper  model. 

9.  Does  a  stool  with  three  legs  always  stand 
firmly  on    a  flat  floor?     One  with   four   legs? 
Give  reasons.     On  what  condition  does  a  stool 
with  four  legs  stand  firmly  on  a  flat  floor  ? 

10.  In  the  figure  the  point  D  is  not  supposed 

to  lie  in  the  same  plane  as  ABC.  Hence,  does  C  lie 
in  the  plane  of  ABDt  Does  B  lie  in  the  plane  of 
ACD1 

11.  How  many   planes    are    determined   by   four 
points  which  do  not  all  lie  in  the  same  plane  ? 

12.  How   many   planes   are   determined   by  four 
lines  which  all  meet  in  a  point,  but  no  three  of  which 
lie  in  the  same  plane  ? 

13.  What  is  the  locus  of  all  points  common  to  two  intersecting 
planes? 

12,  The  demonstrations  and  constructions  of  solid 
geometry  consist  largely  in  applying  the  theorems  already 
known  in  plane  geometry  to  figures  lying  in  various 
planes  determined  by  points  and  lines  in  space,  as  illus- 
trated in  the  preceding  exercises. 

The  next  following  problems  are  typical  in  this  respect. 


6 


SOLID   GEOMETRY. 


THEOREMS   ON  PERPENDICULARS. 

13,  Definitions.     A  line  is  said  to  be  perpendicular  to  a 
plane  if  it  is  perpendicular  to  every  line  of  the  plane  pass- 
ing through  the  point  in  which  it  meets  the  plane.     In 
this  case  the  plane  is  also  perpendicular  to  the  line.     The 
point  in  which  the  perpendicular  meets  the  plane  is  the 
foot  of  the  perpendicular. 

NOTE.  It  is  obvious  that  at  a  point  in 
a  line  there  may  be  many  lines  in  space  which 
are  perpendicular  to  it.  For  instance,  if  AP 
is  JL  BC,  rotate  AP  about  EC  as  an  axis, 
keeping  ZPAC  =  ZPAB.  Then  AP  re- 
mains _L  BC  in  every  one  of  its  positions. 

Thus,  all  spokes  of  a  wheel  are  perpendic- 
ular to  the  axle. 

14,  A  line  or  plane  is  said  to  be  constructed  whenever 
the  points  which  determine  it  are  constructed. 

15,  PKOBLEM.     Through  a  given  point  to  construct 
a  plane  perpendicular  to  a  given 

line. 

Given :  the  line  /  and  the  point  P. 

To  construct    a   plane   through  p      /     p 
perpendicular  to  I. 

ffjf 

Construction,     (#)    Let   P    be   on 

the  given  line  1.  Through  P  construct  two  lines  Zx  and 
?2,  each  perpendicular  to  I.  Then  the  plane  determined 
by  Zj  and  12  is  the  required  plane. 

Proof  :  Connect  B  and  C,  any  two  points  different  from 
P  on  Zj  and  Z2  respectively. 

Through  P  draw  any  line  Z8  meeting  BC  in  D. 

On  I  lay  off  PE  =  PF  and  complete  the  figure  as  shown. 


E 


D 


LINES    AND    PLANES  IN  SPACE. 


Now  prove    (1)  A  EEC  ^  A  FBC,    (2)  Z  EBD  =  Z 
(3)  A  #BD  ^  A  .FjBD,    (4)   A  EPD  ^  A  .PPD,     (5)   Z 
Z  .FPD.     Hence,  PD  -L  I. 

This  proof  holds  for  every  such  line  13  except  the  line 
through  P  parallel  to  BC.  In  this  case  we  can  select  an- 
other point  Cf  on  12  so  that  13  shall  not  be  parallel  to  BCr. 

Since  PD  is  any  line  through  P  in  the  plane  of  ^  and  ?2, 
it  follows  by  definition  that  the  plane  determined  by  ^ 
and  12  is  perpendicular  to  I. 

(5)  If  the  point  P  is  not  on  the  line  I,  draw  a  line  from 
P  perpendicular  to  I  at  some  point  Q,  and  then  as  above 
construct  a  plane  through  Q  perpendicular  to  I. 


16,  EXERCISE. 

Prove  that  through  a  point  there  is  not  more  than  one  plane  per- 
pendicular to  a  given  line. 

i 

17,  PROBLEM.     At  a  point  in  a  plane  to  construct  a 
line  perpendicular  to  the  plane. 

Given :  the  point  P  in  the  plane  M. 

To  construct  a  line  perpendicu- 
lar to  M  at  P. 

Construction.   Let  ^  be  any  line 
in  M  through  P. 

Pass  a  plane  N  through  P  J_  ^ 

(§  15). 

Let  plane  N  cut  plane  M  in  ?2  (§  10). 

In  plane  N  erect  PA  _L  12. 

Then  PA  is  the  perpendicular  required. 

Proof  :  Show  that  PA  is  _L  to  both  l±  and  12  and  hence  to 
their  plane,  that  is,  to  the  plane  M. 


SOLID   GEOMETRY. 


18.  EXERCISES. 

1.  At  a  point  in  a  plane  only  one  line  can  be  erected  perpendicu- 
lar to  the  plane. 

PROOF.  Suppose  a  second  perpendicular  is  possible.  Then  we  shall 
have  two  lines  as  /3  and  14  perpendicular  to  M  at  P.  Let  the  plane  of 
ls  and  14  intersect  M  in  a  line  Z5.  Then  /3  and  14  are  both  JL  to  15  and 
lie  in  the  same  plane  with  it,  which  is  impossible. 

2.  Take  two  pieces  of  cardboard  in  one  of  which  a  line  is  drawn 
at  random,  and  in  the  other  a  line  is  drawn  perpendicular  to  an  edge. 
Place  them  so  as  to  illustrate  the  construction  in  §  17. 


19,  PROBLEM.  From  a  point  outside  a  plane  to  con- 
struct a  line  perpendicular  to  the  plane. 

Given :  the  plane  M  and  the  point  P  outside  it. 
To  construct  a  line  from  P_L  M. 

Construction.     Let   I   be    any 
line  in  plane  M. 

From  P  draw  PA  _L  I,  and  in 
plane  M  draw  AE  J_  I  at  A. 

Finally,      draw      PO  _L  AK. 
Then  PO  is    the    required  line  j/7 

perpendicular  to  M. 

Proof :  Through  O  in  M  draw  any  other  line  OB  meet- 
ing I. 

Prolong  PO  to  P',  making  OP'  =  OP,  and  connect  points 
as  shown  in  the  figure. 

Then  line  I  J_  plane  P^lPr  (why?). 

Now  show  that  (1)  PA  =  PfA,  (2)  A  APE  3*  A  AP'B, 
(3)  A  BOP  ^  A  BOP'. 

Hence,  PO  -L  M.     (Why  ?) 


LINES  AND  PLANES  IN   SPACE.  9 

20  EXERCISES. 

1.  Only  one  perpendicular  to  a  plane  can  be  drawn  from  a  point 
outside  the  plane. 

SUGGESTION.     Show  that  the  hypothesis  of  two  perpendiculars  from 
an  outside  point  to  a  plane  leads  to  a  contradiction. 

2.  A  perpendicular  is  the  shortest  distance  from  a  point  to  a  plane. 

3.  A  line  cannot  be  perpendicular  to  each  of  two   intersecting 
planes. 

SUGGESTIONS.  (1)  If  a  line  I  is  perpendicular  to  the  planes  M  and  N 
at  the  points  A  and  B,  and  C  is 
a  point  in  their  intersection, 
then  A  ABC  would  contain 
two  right  angles.  (2)  If  I  is 
perpendicular  to  M  and  N  at 
a  point  P  in  their  intersec- 
tion, pass  a  plane  through  I,  meeting  M  and  N  in  l}  and  12. 

4.  A  line  which  is  perpendicular  to  each  of  two  lines  at  their  point 
of  intersection  is  perpendicular  to  their  plane.     See  §  15. 

5.  All  perpendiculars  to  a  line  at  a  point  in  it  lie  in  the  same 
plane,  namely,  the  plane  determined  by  any  two  of  them. 

The  following  are  theorems  proved  in  the  preceding  con- 
structions and  exercises. 

21,  THEOREM.      Through  any  given  point   there   is 
one  and  only  one  plane  perpendicular  to  a  given  line. 

22,  THEOREM.     Through  any  given  point  there  is  one 
and  only  one  line  perpendicular  to  a  given  plane. 

23,  THEOREM.     All  lines  perpendicular  to  a  line  at  a 
point  lie  in  one  and  the  same  plane. 

24,  THEOREM.     A  line  perpendicular  to  each  of  two 
intersecting  lines  is  perpendicular  to  their  plane. 

25,  THEOREM.     The  shortest  distance  from  a  point 
to  a  plane  is  the  perpendicular  to  the  plane. 


10  SOLID   GEOMETRY. 

26,  EXERCISES. 

1.  Show  how  a  carpenter  could  use  the  theorem  of  §  24  to  stand 
a  post  perpendicular  to  the  floor,  by  use  of  two  ordinary  steel  squares. 

2.  If  a  steel  square  be  made  to  rotate  about  one  of  its  edges  as  an 
axis,  what  kind  of  surface  does  the  other  edge  describe  ? 

3.  In  making  long   timbers  perpendicular  to  each  other,  a  car- 
penter may  use  simply  a  measuring   rod,    such   as   a  ten-foot  pole. 
Show  how  to  make  a  timber  perpendicular  to  the  floor  by  this  process. 

4.  Show  how  a  back-stop  on  a  ball  field  can  be  adjusted  perpen- 
dicular to    the  line  through  second 

base  and  the  home  plate.  What 
theorems  of  solid  geometry  are 
used? 

5.  If    a   plane   is   perpendicular 
to  a  line-segment  PP'  at  its  middle 
point,   prove:    (1)   Every  point  in 
the  plane  is  equally  distant  from  P 
and  P' ;  (2)  every  point  equally  dis- 
tant  from   P  and  P'  lies   in    this 

plane.     What  is  the  locus  of  all  points  in  space  equidistant  from  P 
and  P'  ?     See  §  127,  Plane  Geometry. 

6.  Given  the  points  A  and  B  not  in  a  plane  M.     Find  the  locus 
of  all  points  in  M  equidistant  from  A  and  B. 

SUGGESTION.  All  such  points  must  lie  in  the  plane  M  and  also  in 
the  plane  which  is  the  perpendicular  bisector  of  the  segment  AB. 

Is  there  any  position  of  the  points  A  and  B  for  which  there  is  no 
such  locus  ?  For  which  the  locus  contains  the  whole  plane  M ? 

7.  On  a  line  find  a  point  equidistant  from  two  given  points  not  on 
the  line. 

Is  there  any  position  of  the  points  for  which  there  is  no  such  point 
on  the  line  ?  For  which  all  points  on  the  line  satisfy  this  condition  ? 

8.  Find   the   locus  of   all  points  in  space  equidistant  from  the 
three  vertices  of  a  triangle. 

9.  Find  the  locus  of  all  points  in  a  given  plane  equidistant  from 
the  vertices  of  a  triangle  not  in  the  plane. 

Is  there  any  position  of  the  triangle  for  which  this  locus  contains 
more  than  one  point  ?  No  point  ? 


LINES  AND  PLANES  IN   SPACE. 


11 


10.  Ask  and  answer  questions  similar  to  the  two  preceding,  using 
the  vertices  of  a  square  instead  of  a  triangle ;  the  vertices  of  a  rec- 
tangle ;  of  any  polygon  which  can  be  inscribed  in  a  circle. 

11.  Find  the  locus  of  all  points  in  space  equidistant  from  the 
points  of  a  circle  and  also  of  all  points  in  a  plane  equidistant  from  the 
points  of  a  circle  not  in  that  plane.     Discuss  as  in  Ex.  9. 

12.  Find  the  locus  of  all  points  equidistant  from  two  given  points 
A  and  B,  and  also  equidistant  from  two  points  C  and  D.     Discuss. 

13.  In  geometry  of  two  dimensions,  how  many  lines  may  be  per- 
pendicular to  a  given  line  at  a  given  point?    In  geometry  of  three 
dimensions? 

14.  In  how  many  points  can  a  straight  line  cut  a  plane  ?    In  how 
many  points  may  it  cut  a  curved  surface,  such  as  a  stovepipe?    In 
how  many  points  can  a  straight  line  cut  a  circle?    In  how  many 
points  can  a  plane  cut  a  circle,  the  plane  being  distinct  from  the 
plane  of  the  circle? 

15.  State  and  prove  a  theorem  of  solid  geome- 
try corresponding  to  the  theorem  in  the  Plane 
Geometry,  comparing  the  lengths  of  oblique  seg- 
ments cutting  off  equal  distances  from  the  foot 
of  a  perpendicular. 

Also  state  and  prove  the  converse. 


State   and  prove   a  theorem  comparing  the 


lengths  of  segments  cutting  off  unequal  distances.     See  §§  112,  115, 
116,  Plane  Geometry. 

16.  Find  the  locus  of  all  points  in  a  plane  which  are  equally  dis- 
tant from  a  given  point  outside  the  plane.     If  a  perpendicular  be 
drawn  to  the  plane  from  this  point,  how  is  its  foot  related  to  this 
locus  ? 

17.  If  in  the  figure  PD  JL  plane  M,  and 
DC  -L  AB  any  line  of  the  plane,  prove  that 
PC  ±AB. 

SUGGESTION.      Lay   off    CA  =  CB,  and 
compare  triangles. 

18.  If  in  the  same  figure  PD  _L  M,  and 
PC  ±AB  any  line  of  the  plane,  prove  that 
DC'±AB. 


12  SOLID   GEOMETRY. 

THEOREMS   ON  PARALLELS. 

27.  THEOREM.     If  tivo  lines  are  perpendicular  to 
the  same  plane,  they  are  parallel. 

CONVERSELY.    If  one  of  two  parallel  lines  is  perpen- 
dicular to  a  plane,  the  other  is  also. 

Given :  (1)  AB  and  CD  each  _L  to 
the  plane  M. 

To  prove  that  AB  II  CD. 

Proof:  Draw  BD  and  make 
DE  J.  DB. 

Take  points  A  and  E  so  that 
BA  =  DE,  and  draw  AD,  AE,  and 
BE. 

Now  prove  (1)  A  ABD  ^  A  BDE.       (2)  A  ADE  ^  A  ABE. 

Hence,  Z  ^IDJ£  is  a  right  angle,  and  DC,  IM,  and  DB  all 
lie  in  the  same  plane.  (Why?) 

Therefore  CD  and  AB  are  in  the  same  plane  and  perpen- 
dicular to  the  same  line  BD.     (Why?) 
Hence,  AB  \\  CD. 

Given :  (2)  AB  II  CD  and  CD  _L  M. 

To  prove  that  AB  J_  if. 

Proof:  If  AB  is  not  _L  M,  suppose  A'B  _L  if.  Then 
A'B  II  CD.  But  through  B  there  is  only  one  line  li  CD. 

Hence,  AfB  and  AB  coincide. 

But  AfB  was  taken  _L  M.     Hence,  AB  J_  M. 

HISTORICAL  XOTE.  The  above  proof  for  the  direct  case  is  the  one 
given  by  Euclid.  Many  proofs  have  been  given,  but  this  seems  the 
most  elegant. 

28.  COROLLARY.     Two  lines  in  space,  each  parallel 
to  the  same  line,  are  parallel  to  each  other. 


LINES  AND  PLANES  IN  SPACE.  13 

SUGGESTIONS.  Let  ^  II  1B  and  Z2  II  13.  To  prove  7X  II  ?2. 
Take  a  plane  Jf  -L  1B  and  finish  the  proof. 

29,  Definitions.     Two  planes  which  do  not  meet  are  said 
to  be  parallel. 

A  straight  line  and  a  plane  which  do  not  meet  are  said 
to  be  parallel. 

30,  THEOREM.     If  each  of  two  planes  is  perpen- 
dicular to  the  same  line,  they  are  parallel. 

CONVERSELY.  If  one  of  two  parallel  planes  is  per- 
pendicular to  a  line,  the  other  is  also. 

Given:    (1)  M  _L  AB  and  N  J_  AB. 

To  prove  that  M  II  N. 

Proof :  Suppose  M  and  2V  to  meet  in 
some  point  P,  and  show  that  this  leads 
to  a  contradiction.  ^  f 

Given:    (2)  M  II  2V  and  M  JL  AB. 

To  prove  that  N  J_  AB. 

Proof :  Through  AB  pass  a  plane  cutting  M  in  BD  and 
2V  in  AC,  and  also  a  second  plane  cutting  M  in  BF  and  IV  in 
AE. 

Then  JBD  II  AC,  for  if  they  could  meet,  then  M  and  2V 
would  meet,  which  is  contrary  to  the  hypothesis. 

Likewise  BF  II  AE. 

Complete  the  proof,  showing  that  AB  _L  AC,  and  AB  _L  ^2?, 
and  hence  AB  J_  2V. 

31,  COROLLARY.     If  a  plane  intersects  two  parallel 
planes,  the  lines  of  intersection  are  parallel. 


14  SOLID   GEOMETRY. 

32,  THEOREM.     If  a  straight  line  is  parallel  to   a 
given  plane,  it  is  parallel  to  the  intersection  of  any 
plane  through  it  with  the  given  plane. 

SUGGESTION.  If  Zt  is  the  given  line,  and  l-2  the  intersection  of  the 
plane  through  ^  with  the  given  plane,  show  that  ZL  and  12  lie  in  the 
same  plane  and  cannot  meet. 

33,  COROLLARY.     If  a  line  outside  a  plane  is  parallel 
to  some  line  in  the  plane,  then  the  first  line  is  parallel 
to  the  plane. 

34,  THEOREM.     If  two  intersecting  lines  in  one  plane 
are  parallel,  respectively,  to  two  intersecting  lines  in 
another  plane,  then  the  two  planes  are  parallel,  and  the 
corresponding  angles  formed  by  the  lines  are  equal. 


Given:  the  planes  M  and  N  in  which  AB  II  A'B',  and  AC  II  A'C'. 

To  prove  that  M  II  N  and  /.\  =  Z.Z. 

Outline  of  proof  :  (1)  To  prove  M  II  N. 

If  M  and  N  could  meet,  let  I  be  their  line  of  intersection. 
Then  neither  AB  nor  AC  can  meet  I  since  each  is  II  N  by 
§  33.  Hence  AB  II  I  and  AC  II  Z,  which  is  impossible. 
(Why?) 

(2)   To  prove  Z.  1  =  Z.  2,  study  the  following  analysis : 

Lay  off  AB  =  ArBr,  AC=ArCf,  and  draw  BC,  BfCf,  AAf, 
BBf,  and  CCr . 

Then  Z1  =  Z2  if  A  ABC^A  AfB!Cr,  which  is  true  if 
BC=B!Cr.  But  BC  =  BrCr  if  BCC'B'  is  a  O,  which  is  so 
if  BB1  —  CCf  and  BBf  II  CC1.  This  last  is  true  if  ABBrAf 


LINES  AND  PLANES  IN  SPACE.  15 

and  ACC'A'  are  O7,  for  then  ££'  =  CC1  =  ^^'  and  £J3'  II 
AA'  II  CC'. 

Now  reverse  the  order  of  these  steps  and  give  the  proof 
in  full,  with  all  the  reasons. 

35,  COROLLARY.  If  two  angles  in  space  have  their 
sides  respectively  parallel,  the  angles  are  either  equal  or 
supplementary. 

State  this  in  full  detail  as  in  §§  106-108,  Plane  Geometry,  and 
show  how  it  applies  to  the  above  figure. 

36,  EXERCISES. 

1.  Show  that   parallel    line-segments   included  between   parallel 
planes   are   equal,  and   hence   that  parallel  planes  are  everywhere 
equally  distant. 

2.  Find  the  locus  of  all  points  at  a  given  distance  from  a  given 
plane ;  also  of  all  points  equally  distant  from  two  parallel  planes. 

3.  Show  how  to  determine  a  plane  parallel  to  a  given  plane  and 
at  a  given  distance  from  it.     How  would  you  place  three  shelves 
parallel  to  each  other  and  a  foot  apart? 

4.  Show  that  two  straight  lines  in  space  may  not  meet  and  yet 
not  be  parallel. 

5.  Show  that  through  any  given   line  a  plane  may  be    passed 
parallel  to  any  other  given  line  in  space. 

SUGGESTION.  If  l^  and  Z2  are  the  given  lines,  through  any  point 
in  lv  draw  13  II 12,  and  show  that  the  plane  determined  by  Zx  and  ls  is 
the  required  plane  II Z2. 

6.  Through  a  given  point  pass  a  plane  parallel  to  each  of  two 
given  lines  in  space. 

SUGGESTION.  Through  the  given  point  pass  lines  II  to  each  of  the 
given  lines,  then  use  §  33. 

7.  Show  that  through   a  point  outside  a  plane  any  number  of 
lines  can  be  drawn  parallel  to  the  plane.     How  are  all  these  parallels 
situated  ? 

8.  Find  the  locus  of  all  lines  through  a  fixed  point  parallel  to  a 
fixed  plane. 


16  SOLID  GEOMETRY. 

37.  THEOREM.  If  two  straight  lines  are  cut  "by  three 
parallel  planes,  the  intercepted  segments  on  one  line  are 
in  the  same  ratio  as  the  corresponding  segments  on  the 
other. 

,~c       M? 

^V"     \          / 


I         \\  7 

— — Ap  / 


Given :  the  lines  AB  and  CD  cut  by  the  planes  M,  N,  and  P, 

To  prove  that—  =  — . 
^s      GD 

Proof :  Connect  the  points  A  and  J>,  and  let  the  plane 
determined  by  AD  and  CD  cut  the  planes  M  and  jr  in  AC 
and  FG  respectively.  And  let  the  plane  of  AB  and  AD 
cut  2V  and  P  in  EF  and  BZ>  respectively. 

Now  show  that  ^cil-FG  and  EF\\  BZ),  and  hence  complete 
the  proof,  giving  all  the  reasons. 

38.  COROLLARY.  Parallel  planes  which  intercept 
equal  segments  on  any  transversal  line  intercept  equal 
segments  on  every  transversal  line. 

39.  EXERCISES. 

1.  Show  how  to  find  all  the  points  on  the  floor  of  the  schoolroom 
which  are  equally  distant  from  one  of  the  lower  corners  of  the  win- 
dow sill  and  one  of  the  upper  corners  of  the  opposite  door. 

2.  If  the  spaces  between  four  shelves  are  5,  8,  and  10  inches  re- 
spectively, and  a  slanting  rod  intersecting  them  has  a  7-inch  segment 
between  the  first  two  shelves,  find  the  other  two  segments  of  the  rod. 


LINES  AND  PLANES  IN  SPACE.  17 

3.  If  we  attempt  to  stand  up  a  ten-foot  pole  in  a  room  eight  feet 
high,  find  the  locus  of  the  foot  of  the  pole  on  the  floor  when  the  top 
is  kept  at  a  fixed  point  on  the  ceiling. 

4.  Show  that,  if  three  line-segments  not  in  the  same  plane  are 
equal  and  parallel,  the  triangles  formed  by  joining  their  extremities,  as 
in  the  figure  of  §  34,  are  congruent,  and  their  planes  are  parallel. 

5.  Given  a  plane  M  and  a  point  P  not  in  M.     Find  the  locus  of 
the  middle  points  of  all  segments  connecting  P  with  points  in  M. 

6.  Given  a  plane  M  and  a  point  P  not  in  M.     Find  the  locus  of 
a  point  which  divides  in  a  given  ratio  each  segment  connecting  P  with 
a  point  in  M :    (a)  if  the  segments  are  divided  internally;    (5)  if 
they  are  divided  externally. 

7.  The  locus  required  in  Ex.  6  consists  of  two  planes,  each  parallel 
to  the  given  plane  M.     Are  these  two  planes  equally  distant  from  M ? 

8.  Show  that  a  plane  containing  one  only  of  two  parallel  lines  is 
parallel  to  the  other. 

9.  If  in  two  intersecting  planes  a  line  of  one  -is  parallel  to  a  line 
of  the  other,  then  each  of  these  lines  is  parallel  to  the  line  of  intersec- 
tion of  the  planes. 

10.  Show  that  three  lines  which  are  not  concurrent  must  all  lie  in 
the  same  plane,  if  each  intersects  the  other  two. 

11.  Show  that  three  planes,  each  of  which  intersects  the  other 
two,  have  a  point  in  common  unless  their  three  lines  of  intersection 
are  parallel. 

SUGGESTION.  Suppose  two  of  the  intersection  lines  are  not  parallel, 
and  meet  in  some  point  O.  Then  show  that  the  other  line  of  inter- 
section passes  through  0,  and  hence  that  0  is  the  point  common  to  all 
three  planes. 

12.  Given  two  intersecting  planes  M  and  N.     Find  the  locus  of 
all  points  in  M  at  a  given  distance  from  N. 

13.  Given  two  non-intersecting  lines  Zx  and  Z2.     Find  the  locus  of 
all  lines  meeting  /t  and  parallel  to  la. 

14.  Prove  that  the  middle  points  of  the  sides  of  any  quadrilateral 
in  space  are  the  vertices  of  a  parallelogram. 

SUGGESTION.  Use  the  fact  that  a  line  bisecting  two  sides  of  a  tri- 
angle is  parallel  to  the  third  side.  Note  that  the  four  vertices  of  a 
quadrilateral  in  space  do  not  necessarily  all  lie  in  the  same  plane. 


18 


SOLID   GEOMETRY. 


DIHEDRAL   ANGLES. 

40,  Definitions.     The  part  of  a  plane  on  one  side  of  a 
line  in   it  is  called  a  half-plane.     The  line  is  called  the 
edge  of  the  half -plane.     Two   half -planes   meeting  in  a 
common  edge  form  a  dihedral  angle.     The  common  edge 
is  the  edge  of  the  angle  and  the  half -planes  are  its  faces. 

Lines  in  the  faces  of  a  dihedral  angle  perpendicular  to 
its  edge  at  a  common  point  form  a  plane 
angle,  which  is  called  the  plane  angle  of  the 
dihedral  angle. 

Thus,  in  the  figure,  the  half-planes  M  and  N  D 
have  the  common  edge  AB,  and  form  the  dihedral 
angle  M-AB-N,  read  by  naming  the  two  faces 
and  the  edge.  The  Z  CDE,  whose  sides  are  CD  _L 
AB  in  N  and  ED  ±AB  in  M  is  the  plane  angle  of 
the  dihedral  angle  M-AB-N. 

41,  By  §  34  all  plane  angles  of  a  dihedral 
angle  are  equal  to  each  other. 

A  dihedral  angle  may  be  thought  of  as 
generated  by  the  rotation  of  a  half-plane 
about  its  edge.  The  magnitude  of  the 
angle  depends  solely  upon  the  amount  of 
rotation. 

42,  Two  dihedral  angles  are  equal  when  they  can  be  so 
placed  that  their  faces  coincide. 

43,  THEOREM.     Two  dihedral  angles  are  equal  if 
their  plane  angles  are  equal.  A^ -^     A' 

Given:   the  dihedral  angles  M- 
AB-N  and  M'-A'B'-N1  in  which  the    D 
plane  A  CDE  and  C'D'E'  are  equal. 

To  prove  that 

M-AB-N  =  M'-A'B'-N'. 
Proof :  Place  the  equal  A  CDE  and  C'D'E'  in  coincidence. 


LINES  AND  PLANES  IN  SPACE. 


19 


Then  the  edges  AB  and  A'B'  must  also  coincide,  since 
they  are  both  perpendicular  to  the  plane  CDE  at  the  point 
D  (§  22). 

Face  M  then  coincides  with  M\  since  its  determining  lines, 
AB  and  DE,  coincide  respectively  with  those  of  Mf,  namely, 
AfBf  and  DfE'. 

Likewise,  face  N  coincides  with  Nf. 

Hence,  M-AB-N  =  M'-A'B'-C'  since  their  faces  have  been 
made  to  coincide  (§  42). 

44.  THEOREM.     State  and  prove  the  converse  of  the 
preceding  theorem. 

45.  Definitions.     It  follows  from  §§  43  and  44  that  the 
plane  angle  of  a  dihedral  angle  may  be  regarded  as  its 
measure. 

A  dihedral  angle  is  right,  acute,  or  obtuse  according  as 
its  plane  angle  is  right,  acute,  or  obtuse. 

Two  dihedral  angles  are  adjacent, 
vertical,  supplementary,  or  comple- 
mentary, according  as  their  corre- 
sponding plan  e  angles,  with  a  common 
vertex,  are  adjacent,  vertical,  supple- 
mentary, or  complementary. 

In  the  figure  pick  out  all  the  dihedral 
angles  and  describe  them  and  their  relations,  (1)  if  Z.  CDE  is  acute, 
(2)  if  £  CDE  is  a  right  angle. 


46. 


EXERCISES. 


1.  State  theorems  on  dihedral  angles  corresponding  to  those  on 
plane  angles  in  §§  64,  68,  69,  72-76,  Plane  Geometry. 

2.  State  theorems  concerning  two  planes  cut  by  a  transversal  plane 
corresponding  to  those   on    lines   in   §§  90,   92,    93,  97,   99,   Plane 
Geometry. 

Note  that  in  Exs.  1  and  2  the  proofs  are  exactly  analogous  to  those 
in  the  Plane  Geometry. 


20 


SOLID   GEOMETRY. 


47,  Definition.      Two  planes  are  perpendicular  to  each 
other  if  their  dihedral  angle  is  a  right  angle. 

48,  THEOREM.     If  a  line  is  perpendicular  to  a  plane, 
every  plane  containing  this  line  is  perpendicular  to  the 
plane. 


In  N 


Given :  a  line  /  _L  plane  N  at  P, 

To  prove  that  a  plane  M  containing  I  is  _L  N. 

Proof :  Let  l±  be  the  intersection  of  N  and  M. 

aw  12  -L  Zj  at  P. 

Then  I  JL  lv  12  J-  lv  and  Z2  _L  Z.  (Why?) 

Hence,  JtfJ.AT.  (Why?) 


49. 


EXERCISES. 


1.  Show  that  the  above  theorem  is  equivalent  to  the  following  > 
If  a  plane  is  perpendicular  to  a  line  lying  in  another  plane,  then  the 
first  plane  is  perpendicular  to  the  second. 

2.  Name  all  the  dihedral  angles  in  the  accom- 
panying figure. 

3.  Find  the  locus   of   all   points   equidistant 
from  two  given  parallel  planes  and  also   equi- 
distant from  two  given  points. 

4.  Find  the  locus  of  all  points  at  a  given  dis- 
tance from  a  given  plane  and  equidistant  from  two  given  points. 

Discuss  Exs.  3  and  4  for  the  various  cases  possible. 

5.  How  can  the  theorem  of  §  48  be  used  to  erect  a  plane  perpen- 
dicular to  a  given  plane  ? 


LINES  AND  PLANES  IN  SPACE. 


21 


50,  THEOREM.  If  two  planes  M  and  N  are  mutually 
perpendicular,  and  if  P  is  any  point  in  M,  then  a  line 
through  P  perpendicular  to  N  lies  wholly  in  M  and  is 
perpendicular  to  the  line  of  intersection  of  M  and  N. 

A 


Given  :  plane  M  _1_  plane  N,  and  P  any  point  in  M.  Let  7j  be  the 
intersection  of  M  and  N  and  let  /  be  a  line  through  P_L  N. 

To  prove   that  I  lies  wholly  in  M  and  that  I  _L  lv 
Proof  :  (1)  Let  P  be  any  point  in  M  outside  of  /x. 

Suppose  I  does  not  lie  wholly  in  M. 

Then  through  P  draw  a  line  V  in  the  plane  M  perpen- 
dicular to  Zj  at  O.  Through  o  draw  line  ?2  in  N  _L  lv  Then 
Z2  and  V  form  a  right  angle  (§  47).  Hence  V  _L  N  (§  24). 

But  from  P  there  can  be  but  one  perpendicular  to  N. 

Hence,  I  coincides  with  £',  and  we  have  I  lying  wholly 
in  M  and  I  _L  lr 

(2)  Let  the  point  P  be  in  the  intersection  /j. 

The  proof  is  similar  to  case  (1).     Give  it  in  full. 


51, 


EXERCISES. 


1.  Show  that  if  in  the  figure  of  §  50,  M  ±  N  and  I  is  a  line  in  M 
such  that  I JL  /i,  then  I  ±  N. 

2.  Through  a  given  point,  on  or  outside  of   a  given  plane,  how 
many  planes  can  be  constructed  perpendicular  to  the  given  plane? 


22 


SOLID   GEOMETRY 


52,  THEOREM.  If  a  plane  is  perpendicular  to  each 
of  two  planes,  it  is  perpendicular  to  their  line  of  inter- 
section. 

af 


Given:  M _L  Q  and  N±  Q  and  /  the  intersection  of  M  and  N. 

To  prove  that  Q  JL  I. 

Proof:  From  a  point  P  common  to  M  and  N  draw  a  line 

j-Q. 

Then  Z'  lies  wholly  in  both  M  and  jv.     (Why  ?) 
Hence,  I'  and  Z  are  the  same  line. 
That  is,  I  _L  Q,  or  Q  _L  Z. 


53, 


EXERCISES. 


1.  Any  plane  _L  to  the  edge  of  a  dihedral  angle  is  _L  to  each  of  its 
faces. 

2.  If  three  lines  are  JL  to  each  other  at  a  common  point,  then  each 
is  -L  to  the  plane  of  the  other  two. 

3.  Through  a  given  line  in  space  pass  a  plane  JL  to  a  given  plane. 
How  many  such  planes  can  be  constructed? 

SUGGESTION.     From  some  point  in  the  given  line  draw  a  line  _L  to 
the  given  plane.     Then  use  §  48. 

4.  What  is  the  answer  to  the  question  in  Ex.  3  in  case  the  given 
line  is  perpendicular  to  the  given  plane  ? 

5.  Through  a  given  point  in  space  pass  a  plane  J_  to  each  of  two 
given.planes.     How  many  such  can  be  passed? 

SUGGESTION.     Consider  the  relation  of  the  required  plane  to  the 
intersection  of  the  two  given  planes. 

6.  What  is  the  answer  to  the  question  in  Ex.  5  in  case  the  given 
point  is  on  the  line  of  intersection  of  the  given  planes  ? 


30./ 

LINES  AND  PLANES  IN  SPACE.  23 

54,  Definition.     A  plane  through  the  edge  of  a  dihedral 
angle  bisects  the  angle  if  it  forms  equal  angles  with  the 
faces. 

55,  THEOREM.      The  locus   of  all  points   in   space 
equally  distant  from  the  faces  of  a  dihedral  angle  is  the 
half-plane  bisecting  the  angle. 


A 

Given :  the  plane  P  bisecting  the  dihedral  Z  M-AB-N. 

To  prove :  (1)  that  any  point  E  in  P  is  equally  dis- 
tant from  M  and  .2V,  and  (2)  that  any  point  which  is 
equally  distant  from  M  and  N  lies  in  P. 

Outline  of  proof :  (1)  Draw  EC  _L  M  and  ED  _L  N.  Then 
the  plane  CED  cuts  Jf,  JV,  and  P  in  OC,  OZ>,  and  OE  re- 
spectively and  is  ±  to  both  Jtf  and  N.  (Why  ?) 

Then  Z  .EOD  is  the  measure,  of  P-AB-N  and  Z  .EOC  is 
the  measure  of  P-AB-M.  (Why?) 

Now  use  A  EOC  and  EOD  to  show  that  EC  =  ED. 

(2)  Take  E  any  point  such  that  EC  =  .ED,  and  let  P  be 
the  plane  through  AB  containing  E.  Now  argue  as  in 
§  125,  Plane  Geometry,  to  show  that  Z  EOD=  Z  EOC,  and 
hence  that  Pf  is  the  same  plane  as  the  bisector  plane  P. 

Give  the  proof  in  full. 

56,  COROLLARY.  If  from  any  point  within  a  dihe- 
dral angle  perpendiculars  are  drawn  to  the  faces,  the 
angle  between  the  perpendiculars  is  the  supplement  of 
the  dihedral  angle. 


24 


SOLID   GEOMETRY. 


57.  Definition.     The  projection  of  a  point  on  a  plane  is 
the  foot  of  the  perpendicular  from  the  point  to  the  plane. 

The  projection  of  any  figure  on  a  plane  is  the  locus  of 
the  projections  of  all  points  of  the  figure  on  the  plane. 

58,  THEOREM.     TJie  projection  of  any  straight  line 
on  a  plane  is  a  straight  line  in  that  plane. 


17 


Outline  of  proof:  Pass  a  plane  through  the  given  line 
and  _L  to  the  given  plane.  Is  this  always  possible  ?  (Why  ?) 
Now  show  that  the  intersection  of  these  two  planes  con- 
tains the  projection  of  every  point  of  the  given  line  upon 
the  given  plane. 

59,  THEOREM.  The  acute  angle  formed  by  a  straight 
line  with  its  own  projection  on  a  plane  is  the  least  angle 
which  it  makes  with  any  line  in  that  plane. 


Outline  of  Proof :  Let  BC  be  the  projection  of  AB,  and 
let  BD  be  any  other  line  in  M  through  B. 
Lay  off  BD  =  BC  and  draw  AD. 
In  the  A  ABD  and  ABC  show  that  Z  ABD  >  Z  yUSC. 

60,  Definition.  The  angle  between  a  plane  and  a  line 
oblique  to  it  is  understood  to  mean  the  acute  angle  be- 
tween the  line  and  its  projection  upon  the  plane. 


LINES  AND  PLANES  IN  SPACE.  25 

61,   PROBLEM.     To  construct  a  common  perpendicular 
to  two  non-parallel  lines  in  space. 


Given :  7X  and  72,  two  non-parallel  lines,  and  also  non-intersecting. 
To  construct  a  line  EC  perpendicular  to  each  of  them. 
Construction :    Through  J.,  any  point  in  £2,  draw  Z3  11  lv 
Let  M  be  the  plane  determined  by  ?2  and  l& 
Through  ^  pass  a  plane  P  J_  M  and  meeting  M  in.  ?4. 
At  B  the  intersection  of  12  and  Z4,  in  the  plane  P,  erect 
.BC-l  Jf. 

Then  Bd  is  the  required  perpendicular. 
Proof  :    Give  the  proof  in  full. 

62,  EXERCISES. 

1.  If  a  line  is  ±  to  a  plane,  show  that  its  projection  is  a  point. 

2.  If  a  line-segment  is  ||  to  a  plane,  show  that  its  projection  is  a 
segment  equal  to  the  given  segment. 

3.  If  a  line-segment  is  oblique  to  a  plane,  show  that  its  projection 
is  less  than  the  given  segment. 

4.  If  a  line-segment  6  in.  long  makes  an  angle  of  30°  with  a  plane, 
find  the  length  of  its  projection  on  the  plane.     If  it  makes  an  angle 
of  45° ;  an  angle  of  60°. 

5.  If  two  parallel  lines  meet  a  plane,  they  make  equal  angles  with 
it.    (Why?)    Is  the  converse  true  ? 

6.  If  a  line  cuts  two  parallel  planes,  it  makes  equal  angles  with 
them.     (Why?)     Is  the  converse  true ? 

7.  If  two  parallel  line-segments  are  oblique  to  a  plane,  their  pro- 
jections on  the  plane  are  in  the  same  ratio  as  the  given  segments. 


26 


SOLID   GEOMETRY. 


POLYHEDRAL  ANGLES. 

63,  Definitions.     Given  a  convex  polygon  and  a  point  P 
not  in  its  plane.     If  a  half-line  I  with  end  point  fixed  at  P 
moves  so  that  it  always  touches  the  polygon  and  is  made 
to  traverse  it  completely,  it  is  said  to  generate  a  convex 
polyhedral  angle. 

The  fixed  point  is  the  vertex  of  the  polyhedral  angle, 
and  the  rays  through  the  ver- 
tices of  the  polygon  are  the  edges 
of  the  polyhedral  angle. 

Any  two  consecutive  edges  de- 
termine a  plane,  and  the  portion 
of  such  a  plane  included  between 
its  edges  is  called  a  face  of  the 
polyhedral  angle. 

The  plane  angles  at  the  vertex 
are  called  the  face  angles  of  the  polyhedral  angle.     A  poly- 
hedral angle  having  three  faces  is  called  a  trihedral  angle. 

64,  THEOKEM.    If  two  trihedral  angles  have  the  three 
face  angles  of  the  one  equal  respectively  to  the  three 
face  angles  of  the  other,  the  dihedral  angles  opposite  the 
equal  face  angles  are  equal. 


Given :   the  trihedral  angles  P  and  P1,  in  which  Z  a  =  Z  a', 
&=Z6',  Zc=Zc'. 


•So  r 

LINES  AND  PLANES  IN  SPACE.  27 

To  prove  that  the  pairs  of  dihedral  angles  are  equal  whose 
edges  are  PA  and  P'A',  PB  and  P'B',  PC  and  p'c'. 

Proof :  Let  p  and  pr  be  cut  by  the  planes  ABC  and 
A'B'C',  making  PA  =  PB  =  PC  =  P'A'  =  P'B'  =  P'C'. 

On  the  edges  PA  and  P'A'  lay  off  AF=  A'F',  and  through 
F&nd.Fr  pass  planes  _L  to  PA  and  P'A'  respectively. 

Let  the  first  plane  cut  AC  in  some  point  E,  and  AB  in 
some  point  J),  and  likewise  the  second  plane  cut  A'c1  and 
A'B'  in  corresponding  points  Ef  arid  D'. 

Why  must  there  be  such  intersection  points?  Must 
they  be  between  A  and  C,  A  and  1?,  Ar  and  (/,  J/  and  Bf 
respectively  ? 

Then  z§  DFE  and  D'F'E'  are  the  measures  of  the  dihedral 
angles  whose  edges  are  AP  and  A'P',  and  these  are  to  be 
proved  equal  to  each  other. 

Now  use  the  pairs  of  face  triangles  APB,  A'P'B',  etc.,  to 
show  that  A  ABC  ^  A  AfBfCf.  Then  show  in  order 

(1)  AADF^AA'D'F'.        (2)  AAEF^AA'E'F'. 

(3)  AADE^AA'D'E'.        (4)  A  DFE  ^  A  D'F'E'. 

In  like  manner  the  other  pairs  of  dihedral  angles  may  be 
proved  equal  to  each  other. 

65.  Two  polyhedral  angles  are  congruent  if  they  can  be 
so  placed  that  their  vertices  and  edges  coincide. 

A  polyhedral  angle  is  read  by  naming  the  vertex  and  one  letter  in 
each  edge,  as  P-ABCDE,  or  by  the  vertex  alone  where  no  ambiguity 
would  arise. 

66,  COROLLARY.     If  two  trihedral  angles  have  their 
face  angles  equal  each  to  each  and  arranged  in  the  same 
order )  they  are  congruent. 

For  by  the  theorem  their  dihedral  angles  are  equal  each  to  each, 
and  if  the  equal  faces  are  arranged  in  the  same  order,  the  equal  parts 
may  be  applied  in  succession,  and  the  trihedral  angles  may  be  made 
to  coincide  throughout. 


28  SOLID   GEOMETRY. 

SUMMARY  OF   CHAPTER  I. 

1.  Describe  the  various  ways  of  determining  a  plane. 

2.  State  the  axioms  used  in  this  chapter. 

3.  State  the  definitions  and  theorems  on  perpendicular  lines  and 
planes. 

4.  State  the  definitions  and  theorems  on  parallel  lines  and  planes. 

5.  Name  some  applications  in  connection  with  perpendiculars 
and  parallels  which  have  impressed  you. 

6.  State  some  facts  in  regard  to  perpendiculars  and  parallels  in 
the  plane  which  do  not  hold  in  space. 

7.  Give  the  definitions  and  theorems  on  dihedral  angles. 

8.  What  theorems  on  perpendicular  planes  are  proved  in  connec- 
tion with  dihedral  angles? 

9.  Give  the  definitions  and  theorems  on  projections  used  in  this 
chapter. 

10.  Give  the  definitions  and  theorems  on  polyhedral  angles  thus 
far  used. 

11.  Make  a  list  of  all  the  loci  problems  in  this  chapter. 

12.  Study  the  following  collection  of  problems  and  applications, 
and  then  make  a  collection  of  those  which  impress  you  as  most  inter- 
esting or  useful.     Include  in  this  list  any  applications  in  the  chapter. 

PROBLEMS   AND   APPLICATIONS. 

1.  A  Christmas  tree  is  made  to  stand  on  a  cross-shaped  base.     If 
the  tree  is  perpendicular  to  each  piece  of  the  cross  is  it  perpendicular 
to  the  floor? 

2.  If  the  projections  on  a  plane  of  a  number  of  points  outside  the 
plane  lie  in  the  same  straight  line,  do  the  points  themselves  neces- 
sarily lie  in  a  straight  line? 

3.  If  A,  B,  and  C  do  not  lie  in  the  same  line,  and  if  their  projections 
on  a  plane  M  do  lie  in  a  straight  line,  what  is  the  relation  of  the 
planes  M  and  ABC1 

4.  Is  it  possible  to  project  a  circle  upon  a  plane  so  that  the  projec- 
tion shall  be  a  straight  line-segment?    If  so,  how  must  the  circle  and 
the  plane  be  related  ? 


LINES  AND  PLANES  IN  SPACE. 


; 

29 


5.  How  many  planes  may  be  made  to  pass  through  a  given  point 
parallel  to  a  given  line  ?    Discuss  the  mutual  relations  of  all  such  planes. 

6.  Through  a  point  P  construct  a  line  meeting  each  of  two  lines 
/x  and  /2. 

SUGGESTION.  Let  M  and  N  be  the  planes  determined  by  P  and  ^ 
and  by  P  and  12.  Is  this  construction  always  possible  ? 

7.  Given  a  plane  M  and  lines  lv  12. 
Construct  a  line  perpendicular  to  M  and 
meeting  both  Zj  and  12. 

SUGGESTION.     Project  Zx  and  Z2  on  .M". 

8.  A  line  I  is  parallel  to  a  plane  M, 
and  lines  l±  and  £2  in  3/  are  not  parallel 
to  I.     Show  that  the  shortest  distance 
between  I  and  Zx  is  equal  to  the  shortest 
distance  between  I  and  12. 

9.  Prove  that  the  planes  bisecting 
the  dihedral  angles  of  a  trihedral  angle 
meet  in  a  line. 

10.  Find  the  locus  of  all  points  equidistant  from  the  planes  deter- 
mined by  the  faces  of  a  trihedral  angle. 

11.  Find  the  locus  of  all  points  equi- 
distant from  the  edges  of  a  trihedral 
angle. 

SUGGESTION.  On  the  edges  lay  off 
PA  =  PB  =  PC.  Let  0  be  equidistant 
from  A,  B,  and  C.  Then  any  point  Q 
in  PO  is  equidistant  from  the  edges. 

12.  Planes  determined  by  the  edges 
of  a  trihedral  angle  and  the  bisectors  of 
the  opposite  face  angles  meet  in  a  line. 

SUGGESTION.  Tf,  in  the  figure  of  Ex.  9,  PA  =  PB  =  PC,  and  if 
PE,  PF,  PG  bisect  the  face  angles,  then  E,  F,  G  are  the  middle 
points  of  the  sides  of  the  triangle  ABC.  Now  apply  the  theorem  that 
the  medians  of  a  triangle  meet  in  a  point. 

13.  Planes  perpendicular  to  the  faces  of  a  trihedral  angle  and 
bisecting  its  face  angles  meet  in  a  line. 


30 


SOLID   GEOMETRY. 


/M 


&> 


14.  Show  how  to  locate  a  point  which  is  at  a  distance  of  2  feet 
from  each  face  of  a  trihedral  angle. 

SUGGESTION.  Pass  a  plane  parallel  to  each  face  of  the  trihedral 
angle  and  at  a  distance  of  2  feet  from  it. 

15.  Show  how  to  locate  a  point  which  is  2  feet  from  one  face  of  a 
trihedral  angle,  3  feet  from  the  second,  and  4  feet 

from  the  third. 

16.  Find  the  locus  of  a  point  in  space  such  that 
the  difference  of  the  squares  of  its  distances  from 
two  fixed  points  is  constant. 

SUGGESTION.  First  solve  the  problem  in  the  plane,  obtaining  as  the 
required  locus  the  line  PP'.  Then  rotate  the  figure  about  the  line  AB 
as  an  axis.  See  Plane  Geometry,  page  227,  Ex.  7. 

17.  Find  the  locus  of  all  points  in  a  plane 
at  which  lines  from  a  fixed  point  P  not  in  the 
plane  meet  the  plane  at  equal  angles. 

18.  Find  the  locus  of  all  points  which  are  at 
the  same  fixed  distance  from  each  of  two  inter- 
secting planes. 

SOLUTION.     Pass  a  plane  perpendicular  to  the  intersection  of  the 
two  given  planes,  making  the  plane  cross 
section  shown  in  the  figure.     Draw  the 
bisector  of  Z  AOB  and  a  line  at  the  re- 
quired distance  a  from  OA  and  parallel  p. 
to  it. 

Then  P  is  at  the  distance  a  from  each 
of  the  lines  OA  and  OB.  In  the  same 
manner  three  other  such  points,  PrP2,  P3> 
are  constructed. 

Now  let  this  figure  move  through  space  parallel  to  itself,  the  point 
0  moving  along  the  intersection  of  the  given  planes.  The  points  P, 
Pv  P2,  P3  will  thus  move  along  straight  lines  which  constitute  the 
required  locus. 

19.  Find  the  locus  of  points  3  feet  from  one  of  two  intersecting 
planes  and  6  feet  from  the  other. 

20.  It  is  required  that  a  series  of  electric  lights  shall   be  7  feet 
above  the  floor  of  a  room  and  3  feet  from  the  walls.     Find  the  locus 
of  all  points  at  which  such  lights  may  be  placed. 


LINES  AND  PLANES  IN  SPACE.  31 

21.  Find  the  locus  of  all  points  in  space  equally  distant  from  each 
of  two  intersecting  straight  lines. 

22.  Find  the  locus  of  all  points  in  space  equally  distant  from  two 
parallel  lines. 

23.  Given  two  points  A  and  B  on  the  same  side  of  a  plane  M. 
Determine  a  point  P  in  M  such  that  AP  +  PB  shall  be  a  minimum. 

SUGGESTION.     Pass  a  plane  through  A  and  B  perpendicular  to  M, 
and  proceed  as  in  Ex.  8,  page  200,  Plane  Geometry. 

24.  Show  that  if  the  edge  of  a  dihedral  angle  is  cut  by  two  parallel 
planes,  the  sections  which  they  make  with  the  faces  form  equal  angles. 

25.  Show  that  if  all  edges  of  a  trihedral  angle  are  cut  by  each  of  a 
series  of  parallel  planes,  the  intersections  form  a  series  of  similar  tri- 
angles. 

26.  Find  the  locus  of  the  intersection  points  of  the  medians  of  the 
triangles  obtained  in  Ex.  25.     Also  of  the  altitudes. 

27.  If  three  planes  are  so  related  that  the  segments  intercepted  on 
any  transversal  line  are  in  the  same  ratio  as  the 

segments  intercepted  on  any  other  transversal      /       ^  ~M~I 

then  the  planes  are  parallel. 

SUGGESTION.     Let  M,   N,  Q  be   the  three 


planes.     From  A  any  point  in  M  draw    three  \     \  " 

lines,  not  in  the  same  plane,  meeting  N  in  A',       I    A^^r-^c//  QJ 
B',  C",  and  Q  in  A",  B",  C".     Use  the  hypothe-       -  ' 
sis   to  show  that  A'B1  II  A"B"  and  C'B'  II  C"B".     Hence  Q  II  N. 
Similarly  show  that  M  II  N. 

28.  A  segment  AB  of  fixed  length  is  free  to  move  so  that  its  end- 
points  lie  in  two  fixed  parallel  planes.     Find  the  locus  of  a  point  C  on 
AB  if  AC  is  of  fixed  length. 

29.  If  the  projections  of  a  set  of  points  on  each  of  two  planes  not 
parallel  to  each  other  lie  in  straight  lines,  show  that  the  points  them- 
selves lie  in  a  straight  line. 

30.  If  Jj  and  12  are  not  parallel  and  non-intersecting,  show  that 
there  is  only  one  plane  through  h  parallel  to  lz. 

31.  Show  that  if  two  lines  are  not  parallel  and  do  not  lie  in  the 
same  plane,  they  have  only  one  common  perpendicular,  and  that  the 
shortest  distance  between  the  lines  is  measured  along  this  perpendic- 
ular. 


CHAPTER   II. 
PRISMS  AND  CYLINDERS. 

67,  Definitions.     Any  portion  of  a  plane  entirely  bounded 
by  line-segments  or  curves  is  called  a  plane-segment. 

E.g.  the  portion  of  the  plane  M  inclosed  by  the 
triangle  ABC  is  the  plane-segment  ABC. 

If  the  boundary  is  composed  entirely  of 
straight  line-segments,   the   inclosed  portion  is  called  a 
polygonal  plane-segment. 

68.  A  polyhedron  is  a  three-dimensional  figure  formed 
by   polygonal    plane-segments   which   entirely   inclose   a 
portion  of  space. 

The  line-segments  which  are  common  to  two 
polygons  are  the  edges  of  the  polyhedron. 

The  plane-segments  inclosed  by  the  edges 
are  the  faces,  and  the  intersections  of  the  edges 
are  the  vertices  of  the  polyhedron.     A  poly- 
hedron is  convex  if  every  section  of  it  made  by  a  plane  is 
a  convex  polygon. 

NOTE.  The  word  polyhedron,  as  here  denned,  means  the  surface 
inclosing  a  portion  of  space  and  not  that  portion  of  space  itself. 
However,  it  is  sometimes  convenient  to  use  the  word  polyhedron  when 
referring  to  the  inclosed  space.  Thus,  we  speak  of  dividing  a  poly- 
hedron into  other  polyhedrons  when,  strictly,  we  mean  that  smaller 
polyhedrons  are  constructed  which  divide  into  parts  the  space  inclosed 
in  the  given  polyhedron. 

In  the  same  manner  the  word  polygon  is  sometimes  used  to  indicate 
the  plane-segment  bounded  by  it,  and  the  word  angle  to  indicate  the 
part  of  the  plane  within  it. 


PRISMS  AND   CYLINDERS. 


33 


Thus,  a  face  of  a  polyhedron  is  sometimes  called  a  polygon,  and  the 
face  of  a  polyhedral  angle  is  sometimes  called  an  angle. 

In  all  cases  the  context  will  clearly  indicate  what  is  meant. 

69,  Definitions.     Given  a  convex  polygon  and  a  straight 
line  not  in  its  plane.     If  the  straight 

line  move  so  as  to  remain  parallel  to 
itself  while  it  always  touches  the 
polygon  and  is  made  to  traverse  it 
completely,  the  line  is  said  to  generate 
a  closed  prismatic  surface. 

The  moving  line  is  called  the  gen- 
erator of  the  surface,  and  the  guiding 
polygon  the  directrix. 

A  cross  section  of  a  prismatic  sur- 
face is  made  by  any  plane  cutting  it  and  not  parallel  to 
the  generator. 

70,  A    prism    is    that 
part  of  a  closed  prismatic 
surface  included  between 
two  parallel  cross  sections, 
together  with  the  inter- 
cepted  plane-segments. 

The  parallel  plane-segments  are  the  bases  of  the  prism, 
and  the  portion  of  the  prismatic  surface  between  the 
bases  is  called  the  lateral  surface  of  the  prism. 

The  lateral  surface  is  composed  of  par- 
allelograms (why  ?),  and  these  are  called 
the  lateral  faces  of  the  prism.  The  sides  of 
these  parallelograms,  not  common  to  the 
bases,  are  the  lateral  edges  of  the  prism. 

A  right  section  of    a  prism  is  made   by 
a  plane  cutting  each  of  its  lateral  edges,  extended  if  neces- 
sary, and  perpendicular  to  then}? 


34  SOLID   GEOMETRY. 

71,  Prisms  are. classified  according  to  the  form  of  their 
right  sections,  as  triangular,  quadrangular,  pentagonal,  hex- 
agonal, etc.     A  regular  prism  is  one  whose  right  section 
is  a  regular  polygon. 

A  prism  is  a  right  prism  if  its  lateral  edges  are  perpen- 
dicular to  its  bases ;  otherwise  it  is  oblique. 

The  altitude  of  a  prism  is  the  perpendicular  distance 
between  its  bases.  The  altitude  of  a  right  prism  is  equal 
to  its  edge. 

The  lateral  area  of  a  prism  is  the  sum  of  the  areas  of  its 
lateral  faces. 

The  total  area  is  the  sum  of  its  lateral  area  and  the  area 
of  its  bases. 

THEOREMS  ON  PRISMS. 

72.  THEOREM.     The  cross  sections  of  a  prism  made 
by  parallel  planes  are  congruent  polygons. 


Given  a  prism  cut  by  two  parallel  planes  forming  the  polygons 
ABCDE  and  A'B'C'D'E'. 

To  prove  that  ABCDE ^  A'B'C'D'E'. 
Outline  of  proof:   (1)  Show  that  AB  =  A'B',  BC  =  B'cr, 
etc.,  by  proving  that  ABB* A1 ,  BCCrBr,  etc.,  are  UJ. 

(2)  Show  that  Z  ABC  =  Z.  ArBfCr,  £  BCD  =  Z  B'C'D',  etc. 

(3)  Hence,  show  that  ABCDE  and  A'B'C'D'E'   can  be 
made  to  coincide. 


3/3 

PRISMS  AND  CYLINDERS.  35 

73.  THEOREM.  The  lateral  area  of  a  prism  is  equal 
to  the  product  of  a  lateral  edge  and  the  perimeter  of  a 
right  section. 


Suggestion.  Show  that  the  lateral  edges  are  mutually 
equal  and  that  the  area  of  each  face  is  the  product  of  a 
lateral  edge  and  one  side  of  the  right-section  polygon. 

Complete  the  proof. 

NOTE.  The  form  of.  statement  in  this  theorem  is  the  usual 
abbreviation  for  the  more  precise  form : 

The  lateral  area  of  a  prism  is  equal  to  the  product  of  the  numerical 
measures  of  a  lateral  edge  and  the  perimeter  of  a  right  section. 

Similar  abbreviations  are  used  throughout  this  text. 

74.  COROLLARY.     The  lateral  area  of  a  right  prism 
is  equal  to  the  product  of  its  altitude  and  the  per- 
imeter of  its  base. 

75,  Definitions.      A  polyhedron  which  is   a  part   of   a 
prism  cut  off  by  a  plane  meeting  all  the  lateral  edges, 
but  not  parallel  to  the  base,  is  called  a  trun- 
cated prism. 

Two  polyhedrons  are  said  to  be  added  when 
they  are  placed  so  that  a  face  of  one  coincides 
with  a  face  of  the  other,  but  otherwise  each 
lies  outside  the  other. 


36 


SOLID   GEOMETRY. 


76,  THEOREM.  Two  prisms  are  congruent  if  three 
faces  having  a  common  vertex  in  the  one  are  congruent 
respectively  to  three  faces  having  a  common  vertex  in 
the  other,  and  similarly  placed. 


5D 

V 

.B        G  Bf        G' 

Given  the  three  faces  meeting  at  B  in  prism  P  congruent  re- 
spectively to  the  three  faces  meeting  at  B1  in  prism  P',  and 
similarly  placed. 

To  prove  that  P  can  be  made  to  coincide  with  p'. 

Outline  of  proof:  Trihedral  angles  B  and  Br  are  con- 
gruent. (Why?) 

Now  apply  the  two  prisms,  making  B  coincide  with  B1, 
and  thus  show  in  detail  that : 

(1)  The  lower  bases  coincide. 

(2)  The  lateral  faces  at  B  and  B!  coincide. 

(3)  The  upper  bases  coincide. 

(4)  All  the  lateral  faces  coincide. 
Hence  P  and  Pr  coincide  throughout. 

77,  COROLLARY.     Tioo  right  jirisms  are  congruent 
if  they  have  congruent  bases  and  equal  altitude. 

78,  COROLLARY.     Two    truncated  prisms   are   con- 
gruent if  the  faces  forming  a  trihedral  angle  of  one  are 
equal  respectively  to  the  corresponding  faces  of  the  other. 


PRISMS  AND   CYLINDERS. 


37 


79,  Definitions.      A   parallelepiped   is   a 
prism  whose  bases,  as  well  as  lateral  faces, 
are  parallelograms. 

A  rectangular  parallelepiped  has  all  its 
faces  rectangles. 

A  cube  is  a  parallelepiped  whose  faces 
are  all  squares. 

80,  THEOREM.    Any  two  opposite  faces 
of  a  parallelopiped  are  parallel  and  con- 
gruent. 


Suggestions.     Consider   the   opposite   faces   ABFE  and 
DCGH. 

(1)  Show  that  the  sides  of  the  angles  ABF  and  DCG 
are  parallel,  and  hence  the  planes  determined  by  them  are 
parallel. 

(2)  Show  that  these  faces  are  congruent. 

In  like  manner  argue  about  any  other  pair  of  opposite 
faces. 


81, 


EXERCISES. 


1.  Can  a  parallelepiped  be  a  right  prism  without  being  a  rectan- 
gular parallelepiped  ?     Illustrate. 

2.  Show  that  in  a  rectangular  parallelepiped  each  edge  is  perpen- 
dicular to  all  the  other  edges  meeting  it. 

3.  Is  it  possible  to  construct  a  prism  which  has  no  right  section 
according  to  the  definition  of  §  70  ? 


38 


SOLID   GEOMETRY. 


II 


4.  Show  that  any  section  of  a  parallelo-  JJ  R 
piped  made  by  a  plane  cutting  four  parallel 

edges  is  a  parallelogram. 

5.  A  section  made  by  a  plane  passed 
through  two  diagonally  opposite  edges  of  a 
parallelepiped  is  a  parallelogram. 

E.g.  the  section  through  DH  and  BF 
in  the  figure  of  Ex.  4. 

6.  Show  that  any  two  of  the  four  diagonals  of  a  parallelepiped 
bisect  each  other. 

E.g.  AG  and  CE  in  the  figure.     Make 
use  of  the  preceding  example. 

7.  Show  that  the  diagonals  of  a  rec- 
tangular  parallelepiped  are    all  equal  to 
each  other. 

8.  Show  that  the  square  on  the  diag- 
onal of  a  rectangular  parallelepiped  is  equal 
to  the  sum  of  the  squares  on  the  sides 
meeting  at  a  vertex  from  which  it  is  drawn. 

E.g.  in  the  figure  of  Ex.  6,  AG'2  =AC2  +  CG?  =  AS*  +  BC2  -!-  CG2. 

9.  Find  the  ratio  of  the  diagonal  to  one  edge  of  a  cube. 

10.  Find  the  edge  of  a  cube  whose  diagonal  is  14  inches.     Find 
the  diagonal  of  a  cube  whose  edge  is  16  inches. 

11.  Find  a  diagonal  of  a  rectangular  parallelepiped  whose  edges 
are  6,  8,  and  10  inches  respectively. 

12.  Are  the  diagonals  of  a  cube  perpendicular  to  each  other? 

13.  If  two  diagonals  of  a  rectangular  parallelepiped  meet  at  right 
angles,  and  if  two  of  its  faces  are  squares,  find  the  ratio  of  the  sides  of 
the  remaining  faces. 

14.  If  two  congruent  right  prisms  whose  bases  are  equilateral  tri- 
angles are  placed  together  so  as  to  form  one  prism  whose  base  is  a 
parallelogram,  compare  the  lateral  area  of  the  prism  so  formed  with 
the  sum  of  the  lateral  areas  of  the  original  prisms. 

15.  A  right  prism  whose  bases  are  regular  hexagons  is  divided  into 
six  prisms  whose  bases  are  equilateral  triangles.     Compare  the  lateral 
area  of  the  original  prism  with  the  sum  of  the  lateral  areas  of  the 
resulting  prisms.  • 


PEISMS  AND   CYLINDEB-S. 


J/7 
39 


16.  If  the  perimeter  of  a  right  section  of  a  prism  is  24  inches  and 
its  altitude  6  inches,  what  is  the  smallest  possible  lateral  area?     What 
can  be  said  about  the  largest  possible  area  of  such  a  prism  ? 

17.  Any  section  of  a  prism  made  by  a  plane  parallel  to  a  lateral 
edge  is  a  parallelogram. 

18.  Show  that  for  every  prism  there  is  at  least  one  set  of  parallel 
planes  which  cut  the  prisrn  in  rectangular  sections.     See  suggestions 
Ex.  7,  p.  49. 


VOLUMES  OP  RECTANGULAR  PARALLELOPIPEDS. 

82,  Thus  far  certain  properties  of   prisms  have  been 
studied,  but  no  attempt  has  been  made  to  measure  the 
space  inclosed  by  such  a  solid.     For  this  purpose  we  con- 
sider first  a  rectangular  parallelepiped. 

83,  Definition.     In  case  each  edge  of  a  rectangular  par- 
allelopiped  is  commensurable  with  a  unit  segment,  the 
number  of  times  which  a  unit  cube  is  contained  in  it  is 
the  numerical  measure  or  the  volume  of  the  parallelepiped. 

84,  In  the  case  just  described  in  the 
definition,    the   volume    is    easily   com- 
puted. 

E.g.  if  in  the  figure  one  edge  A  C  is  4  units, 
and  an  adjoining  edge  AB  is  3  units,  then  a 
cube  as  AK,  whose  edge  is  one  unit,  may  be 
laid  off  4  times  along  A  C  and  a  tier  of  3  •  4  =  12 
such  cubes  will  adjoin  the  face  AD,  while  5 
such  tiers  will  exactly  fill  the  space  within  the  solid.  That  is, 
5  •  3  •  4=  60  is  the  number  of  cubic  units  in  the  solid. 

Again,  if  the  given  dimensions  are  3.4,  2.6,  4.5  decimeters  re- 
spectively, then  unit  cubes,  with  edge  one  decimeter,  cannot  be  made 
to  fill  exactly  the  space  inclosed  by  the  figure,  but  cubes  with  edge 
each  one  centimeter  will  do  so,  giving  34,  26,  and  45  respectively  along 
the  three  edges  of  the  solid. 

Hence  the  volume  is 

34  •  26  .  45  =  39,780  cubic  centimeters,  or  39.78  cubic  decimeters. 


40 


SOLID   GEOMETRY. 


85,  Axiom  IV.     Any  rectangular  parallelepiped  in- 
closes a  definite   volume  which  is  greater   than  that 
of  another,  provided  no  dimension  of  the  first  is  less 
than  the  corresponding  dimension  of  the  second,  and 
at  least  one  dimension  is  greater. 

86.  THEOREM.     The  volume  of  any  rectangular  par- 
allelopiped  is  equal  to  the  product  of  the  numerical 
measures  of  its  linear  dimensions. 

Proof:   CASE  1.     If  each  dimension  is  commensurable  with 
the  unit  segment. 

This  is  the  case  treated  in  §  84. 


CASE  2.  If  two  dimensions  are  commensurable  with  the 
unit  segment  and  the  third  is  not. 

By  Axiom  IV  the  parallelepiped  has  a  definite  volume  F. 

Suppose  this  is  not  equal  to  abc  and  that  v  <  abc. 

Let  d  be  a  number  such  that  V  =  abc'.  Then  c'  <  c. 
On  BD  lay  off  BH  =  cf . 

Divide  the  unit  segment  into  equal  parts,  each  less  than 
HD,  and  lay  off  one  of  these  parts  successively  on  J5D, 
reaching  a  point  K  between  H  and  D.  Denote  the'  length 
of  BK  by  c"  and  pass  a  plane  through  K  parallel  to  ABE. 


J/? 

PRISMS  AND  CYLINDERS.  41 

By  Case  1  the  volume  of  the  parallelepiped  AL  is  abc" . 
By  hypothesis,  V—  abc',  but  abc1  <  «fo",  since  c'  <  c". 
Hence,  F<  abc".  (1) 

But  by  Ax.  IV,  v  >  abc".  (2) 

Hence,  the  assumption  that  F  <  abc  cannot  hold. 
In  the  same  manner  show  that  V  >  abc  cannot  hold. 
Hence,  we  have  V  =  abc. 

CASE  3.  If  one  dimension  is  commensurable  with  the 
unit  segment  and  two  are  not. 

CASE  4.  If  all  three  dimensions  are  incommensurable 
with  the  unit  segment. 

The  proofs  in  these  cases  are  similar  to  the  above. 

Thus,  in  Case  3,  when  b  and  c  are  both  incommensur- 
able with  the  unit  segment,  we  obtain  the  parallelepiped 
AL,  two  of  whose  dimensions  a  and  c"  are  commensur- 
able with  the  unit,  and  hence,  by  Case  2,  its  volume  is 
abcff.  Then  the  argument  is  identical  with  that  given 
before.  Hence,  in  all  cases  V=  abc. 

87,  COROLLARY    1.     The    volume   of  a   rectangular 
parallelepiped  is  equal  to  the  product  of  the  numerical 
measures  of  its  l}ase  and  altitude. 

88,  COROLLARY  2.  If  two  rectangular  parallelepipeds 
have  two  dimensions  respectively  equal  to  each  other, 
their  volumes  are  in  the  same  ratio  as  their  third  di- 
mensions ;  and  if  they  have  one  dimension  the  same  in 
each,  their  volumes  are  in  the  same  ratio  as  the  products 
of  their  other  tivo  dimensions. 

For  if  V  and  V  are  the  volumes,  and  a,  b,  c  and  a',  b',  c'  the  dimen- 
sions, then  ¥  =  a'b'c  =?-ti  I  =  6' and  c  =  c';  or  Z  =  ^±a  c  =  c. 
V1     a'-b'-c'     a'  V     a'-b' 


42  SOLID   GEOMETRY. 

VOLUMES  OF  PRISMS  IN  GENERAL. 

89,  From  the  formula  for  rectangular  parallelepipeds, 

Volume  =  length  x  width  x  altitude,  or  V=  abc 

we  deduce  the  volumes  of  prisms  in  general  by  means  of 
the  principle : 

Two  polyhedrons  are  equal  (that  is,  have  the  same 
volume)  if  they  are  congruent,  or  if  they  can  be  divided 
into  parts  which  are  congruent  in  pairs. 

The  sign  =  between  two  polyhedrons  means  that  they 
are  equal  in  volume.  The  word  equivalent  is  sometimes 
used  to  mean  equal  in  volume. 

90,  THEOREM.     The  volume  of  an  oblique  prism  is 
equal  to  that  of  a  right  prism  having  for  its  base  a 
right  section  of  the  oblique  prism  and  for  its  altitude  a 
lateral  edge  of  the  oblique  prism. 

Given  the  oblique  prism  AD\  with 
FGHJK  a  right  section,  and  P'O'IffK' 
a  right  section  of  the  prism  extended 
so  that  the  edge  AA'  =  KK'. 

To  prove  that  the  oblique  prism 
AD1  has  the  same  volume  as  the 
right  prism  EH1 '. 

Outline   of    proof:      Show    that 

(1)  ABCDE^A'B'C'D'E'  ; 

(2)  ABFK^A'B'F'K'; 

(3)  EAKJ^E'A'K'J*. 

Hence,  by  §  78  AH  and  AfHr  are  congruent. 
Now  the  given  prism  ADr  =  AH  +  KD*  (§  75) 

and  the  right  prism        KH*  =  A'H'  +  KDr. 
Hence,  AD1  =  EH*.     (Why?) 


PRISMS  AND   CYLINDERS. 


43 


91,    THEOREM.     The  volume  of  any  parallelopiped  is 
equal  to  the  product  of  its  base  and  altitude. 


B 


Given  the  oblique  parallelopiped  AK,  with  base  b  and  alt.  h. 
To  prove  that  its  volume  is  equal  to  b  times  h. 

Proof :  Considering  face  AL  as  the  base  of  prism  AK,  pro- 
duce the  four  edges  parallel  to  AB,  and  lay  off  ArBr  =  AB. 

Through  A1  and  B'  erect  planes  _L  to  AB ',  thus  cutting 
off  the  right  prism  A1  Kr  with  A' L1  as  one  base. 

Now  considering  C'L'  as  the  base  of  prism  A'K' ',  produce 
the  four  edges  II  to  C'B'  and  lay  off  C"B"  =  C'Br. 

At  c"  and  B"  erect  planes  _L  to  C'B",  cutting  off  the  right 
prism  AuKn,  which  is  a  rectangular  parallelopiped. 

Then  show  that  (1)  h  =  h' =  h" ;  (2)  6  "=£'  =  &"; 
(3)  prisms  AK,  AfKf,  A"K"  are  equal  (§  90). 

But  prism  A"K"=b".h".  (Why?)  Hence,  prism 
AK=b  •  h. 

Write  out  this  proof  in  full. 


44 


SOLID   GEOMETRY. 


92.  THEOREM.     The  volume  of  a  triangular  prism  is 
equal  to  the  product  of  its  base  and  altitude. 

Given  the  triangular  prism  whose 
base  is  ABC. 

To  prove  that  the  volume  of  this 
prism  is  equal  to  the  area  of  A  ABC 
multiplied  by  the  altitude  h,  that 
is,  by  the  perpendicular  distance 
between  ABC  and  EFG. 

Proof:  Complete  the  HJ  ABCD 
and  EFGH  and  draw  DH.  Now 

show  that  CDHG  and  ADHE  are  ZU,  and  hence  that  BH  is  a 
parallelepiped. 

Let  KLMO  be  a  right  section  of  BH,  and  let  KM  be  the 
line  in  which  the  plane  ACGE  cuts  the  plane  KLMO. 

Now    (1)  KLMO  is  a  O.     Hence,  A  KLM  ^  A  KMO. 

(2)  Volume  of  prism  ABC-F  =  area  A  KLM  times  BF. 

(3)  Volume  of   prism   BH  =  area  O  KLMO  times  £.F. 

(4)  Hence,  prism  ABC-F  =  |  prism  BH. 

(5)  But  prism  BH  =  h  x  area  of  O  ABCD. 

(6)  Hence,   prism   ABC-F  =  h  x  £    area   of   O  ABCD 
=  h  x  area  of  A  ^J5(7. 

Therefore  prism  ABC-F  is  equal  to  the  product  of  its 
base  and  altitude.     Give  reasons  for  each  step. 

Ef 

93,  COROLLARY  1.    The    volume    of 

any  prism  is  equal  to  the  product  of  its 
base  and  altitude. 

For  any  prism  can  be  divided  into  triangu- 
lar prisms  by  planes  passing  through  one  edge 
and  each  of  the  other  non-adjacent  edges. 


PRISMS  AND   CYLINDERS.  45 

94,  COROLLARY  2.    Any  two  prisms  of  equal  alti- 
tudes have  the  same  volumes  if  their  bases  are  equal. 

95,  COROLLARY  3.    The  volumes  of  two  prisms  have 
the  same  ratio  as  their  altitudes  if  their  bases  are  equal ; 
and  the  same  ratio  as  the  areas  of  their  bases  if  their 
altitudes  are  equal. 

96,  EXERCISES. 

1.  The  proposition  that  the  volume  of  any  prism  is  equal  to  the 
product  of  its  base  and  altitude  is  of  great  importance.     What  theo- 
rems of  this  chapter  were  used  directly  or  indirectly  in  proving  it? 

2.  If  the  base  of  a  prism  is  36  square  inches  and  its  altitude  12 
inches,  what  is  its  volume  ? 

3.  If  the  perimeter  of  the  base  of  a  prism  and  the  length  of  a 
lateral  edge  are  known,  is  the  lateral  area  thereby  determined  ? 

4.  If  the  area  of  the  base  and  the  length  of  an  edge  are  known, 
can  the  volume  be  found  ? 

5.  What  dimensions  of  a  prism  must  be  known  in  order  to  deter- 
mine its  lateral  area  by  means  of  the  theorems  of  this  chapter  ?     What 
dimensions  must  be  known  to  determine  its  volume  ? 

6.  Parallel  sections  of  a  closed  prismatic  surface  are  congruent 
polygons.     Prove. 

7.  Find  the  edge  of  a  cube  if  its  total  area  is  equal  numerically 
to  its  volume,  an  inch  being  used  as  a  unit. 

8.  A  side  of  the  base  of  a  regular  right  hexagonal  prism  is  3 
inches.     Find  its  altitude  if  its  volume  is  54  V3  cubic  inches.     What 
is  the  total  area  of  this  prism? 

9.  A  side  of  one  base  of  a  regular  right  triangular  prism  is  equal 
to  the  altitude  of  the  prism.     Find  the  length  of  the  side  if  the  total 
surface  is  numerically  equal  to  the  volume. 

10.  Solve  the  preceding  problem  in  case  the  prism  is  a  regular 
right  hexagonal  prism. 

11.  The  volume  of  a.  regular  right  prism  is  equal  to  the  lateral 
area  multiplied  by  half  the  apothem  of  the  base.     Prove. 


46  SOLID  GEOMETRY. 

CYLINDERS. 

97,  Definitions.      A   surface   no   segment   of   which   is 
plane  is  called  a  curved  surface. 

E.g.  the  surface  of  an  eggshell  or  of  a  stovepipe  is  a  curved  surface. 

A  closed  plane  curve  is  one  which  can  be  traced  continu- 
ously by  a  point  moving  in  a  plane 
so  as  to  return  to  its  original  posi- 
tion without  crossing  its  path. 

A    convex    closed    curve  is    one 
which  can  be  cut  by  a  straight  line  in  only  two  points. 

98,  Given  a  closed  convex  plane  curve  and  a  straight  line 
not  in  its  plane.     If  the  straight  line 

moves  so  as  to  remain  parallel  to  itself, 
while  it  always  touches  the  curve  and  is 
made  to  traverse  it  completely,  the  line 
is  said  to  generate  a  closed  convex  cylin- 
drical surface.  The  moving  line  is  the 
generator,  and  the  guiding  curve  the  di- 
rectrix. The  generator  in  any  one  of 
its  positions  is  an  element  of  the  surface. 
A  cross  section  of  a  cylindrical  sur- 
face is  made  by  a  plane  cutting  all  its  elements. 

99,  A  cylinder  is  that  part  of  a  closed  cylindrical  sur- 
face included  between  two  parallel  cross 

sections,  together  with  the  plane-segments 
thus  intercepted.  The  plane-segments  are 
the  bases  of  the  cylinder,  and  the  part  of 
the  cylindrical  surface  between  the  bases 
is  the  lateral  surface. 

The  portion  of  the  generating  line  in  any 
position  which  is  included  between  the  bases  is  an  element 
of  the  cylinder. 


PRISMS  AND   CYLINDERS.  47 

A  right  section  of  a  cylinder  is  made  by  a  plane  cutting 
each  of  its  elements  at  right  angles. 

A  circular  cylinder  is  one  whose  right  section  is  a  circle. 

The  radius  of  a  circular  cylinder  is  the  radius  of  its 
right  section. 

A  cylinder  whose  bases  and  right  sections  are  circles  is 
called  a  right  circular  cylinder.  A  right  circular  cylinder 
is  called  a  cylinder  of  revolutions,  since  it 
may  be  generated  by  revolving  a  rectangle 
about  one  of  its  sides  as  an  axis.  The  side 
opposite  the  axis  generates  the  lateral  sur- 
face, and  the  sides  adjacent  to  the  axis 
generate  the  bases. 

100,  THEOREM.  If  a  plane  contains  an  element  of  a 
cylinder  and  meets  it  in  one  other  point,  then  it  contains 
another  element  also,  and  the  section  is  a  parallelogram. 


A 

Given  the  cylinder  AC  and  a  plane  containing  the  element  AD 
and  one  other  point  as  B. 

To  prove  that  it  contains  another  element  BG  and  that 
ABCD  is  a  parallelogram. 

Proof :   Through  B  pass  a  line  BC II  AD. 

Then  BC  is  an  element  of  the  cylinder.     (Why  ?) 

Also  BC  lies  in  the  plane  ABD. 

Now  show  that  ABCD  is  a  parallelogram. 

What  is  the  section  ABCD  if  AC  is  a  right  cylinder? 


48 


SOLID   GEOMETRY. 


101.  Definition.     If  a  plane  contains  an  element  of  a 
cylinder  and  no  other  point  of  it,  the  plane  is  said  to  be 
tangent  to  the  cylinder,  and  the  element  is  called  the  ele- 
ment of  contact. 

102,  THEOREM.     The  bases  of  a  cylinder  are  con- 
gruent plane-segments. 


*Given  a  cylinder  with  the  bases  b  and  b'. 

To  prove  that  b  ^  b'. 

Proof  :  Take  any  three  points  A,  5,  C  in  the  rim  of  the 
base  b  and  draw  the  elements  at  these  points,  meeting  the 
base  br  in  D,  E,  F. 

Show  that  A  ABC^ADEF. 

Now,  while  the  elements  AD  and  BE  remain  fixed,  con- 
ceive CF  to  generate  the  cylinder. 

Evidently  A  ABC  ^  A  DEF  in  every  position  of  CF. 

Hence,  if  base  b'  is  applied  to  base  b  with  these  triangles 
coinciding  in  one  position,  they  will  coincide  in  every 
position  corresponding  to  the  moving  generator. 

That  is,  b  ^bf. 

103,  EXERCISES. 

1.  Show  that  right  sections  of  any  cylinder  are  congruent,  and 
any  section  parallel  to  the  base  is  congruent  to  the  base. 


PRISMS  AND  CYLINDERS.  49 

2.  If   two  intersecting  planes  are  each  tangent  to  a  cylinder, 
show  that  their  line  of  intersection  is  parallel  to  an  element  of  the 
cylinder  and  also  parallel  to  the  plane  containing  the  two  elements 
of  contact. 

3.  What  is  the  locus  of  all  points  at  a  perpendicular  distance  of 
2  feet  from  a  given  line  ? 

4.  If  the  section  of  a  cylinder  made  by  every  plane  parallel  to  an 
element  of  it  is  a  rectangle,  what  kind  of  cylinder  is  it  ? 

5.  If  the  radius  r  of  a  right  circular  cylinder  is  equal  to  its  alti- 
tude, find  the  distance  from  the  center  of  the  base  to  a  plane  whose 
intersection  with  the  cylinder  is  a  square. 

6.  Roll  a  sheet  of  paper  so  as  to  form  a  circular  cylinder,  that 
is,  one  whose  right  section  is  a  circle.     Now  determine  by  inspection 
the  shape  of  the  base  if  the  paper  is  cut  so  as  to  let  the  cylinder  stand 
in  an  oblique  position.     Also  deform  the  cylinder  so  as  to  make  the 
oblique  base  circular,  and  then  determine  the  shape  of  the  right 
section. 

7.  Show  that  for  every  cylinder  there  is  at  least  one  set  of  parallel 
planes  which  cut  the  cylinder  in  rectangular  sections. 

SUGGESTION.  Project  an  element  on  the  plane  of  the  base,  and 
draw  lines  in  the  base  at  right  angles  to  this  projection.  Through 
these  lines  pass  planes  parallel  to  an  element. 

8.  Is  a  polygon  circumscribed  about  a  convex  closed  curve  neces- 
sarily a  convex  polygon  ?     See  §  160,  Plane  Geometry. 

9.  Is  a  polygon  inscribed  in  a  convex  closed  curve  necessarily 
convex?     (Note  that  a  broken  line  AB,  BC, 

CD,  DE,  EA  which  cuts  itself  as  shown  in  the 
figure  is  not  here  regarded  as  a  polygon.)  The 
answers  to  Exs.  8  and  9  limit  the  polygons  to 
be  used  under  Ax.  V  to  convex  polygons. 
Hence  it  is  not  necessary  to  state  in  that 
axiom  that  these  polygons  must  be  convex. 

10.  If  polygons  other  than  convex  are  permitted,  is  it  possible  to 
construct  one  within  a  convex  closed  curve  whose  perimeter  is  greater 
than  the  length  of  the  curve  ?  Can  such  polygons  be  inscribed  in  the 
curve  ? 


50  SOLID   GEOMETRY. 

MEASUREMENT  OF   THE   SURFACE   AND  VOLUME  OF   A  CYLINDER. 

104.  Thus  far  in  geometry  the  word  area  has  been  used 
in  connection  with  plane-segments  only.     In  some  cases 
the   computation   of    an   area   has   been   possible   by   an 
approximation  process  only,  as  in  the  case  of  some  rec- 
tangles and  of  the  circle. 

In  the  case  of  any  curved  surface  it  is  evident  that  ap- 
proximate measurement  is  the  only  kind  possible  in  terms 
of  a  plane  area  unit,  since  no  such  unit,  however  small, 
can  be  made  to  coincide  with  a  segment  of  a  curved  sur- 
face. 

Theorems  concerning  the  surface  and  the  volume  of  a 
cylinder  are  based  upon  the  following  definitions  and  as- 
sumptions : 

105.  Axiom  V.     Any  convex  closed  curve  lias  a  defi- 
nite  length  and   incloses   a   definite  area,  which   are 
less  respectively  than  the  perimeter  and  area  of  any 
circumscribed  polygon  and  greater  than  those  of  any 
inscribed  polygon. 

Also  the  perimeter  and  area  of  either  the  inscribed 
or  the  circumscribed  polygon  may  be  made  to  differ 
from  those  of  the  curve  by  as  little  as"  we  please  by 
taking  all  the  sides  sufficiently  small. 

106.  Definitions.     A  prism   is   said   to   be   inscribed  in 
a  cylinder  if  its  lateral  edges  are  elements  of  the  cylin- 
der, and  their  bases   lie   in   the   same 

planes. 

A  prism  is  said  to  be  circumscribed 
about  a  cylinder  if  its  lateral  faces  are 
all  tangent  to  the  cylinder,  and  their 
bases  lie  in  the  same  planes. 


PRISMS  AND   CYLINDERS.  51 

107,  Axiom  VI.     The   lateral  surface   of  a   convex 
cylinder  lias  a  definite  area,  and  the  cylinder  incloses  a 
definite  volume,  which  are  less  respectively  than  those 
of  any  circumscribed  prism  and  greater  than  those  of 
any  inscribed  prism. 

108,  THEOREM.     The  lateral  area  of  a  convex  cylin- 
der is  equal  to   the  product  of  an   element   and   the 
perimeter  of  a  right  section. 

Given  a  c6nvex  cylinder  of  which  L  is  the  lateral  area,  e  an  ele- 
ment, and  p  the  perimeter  of  a  right  section. 

To  prove  that  L  =  ep. 

Proof :  It  will  be  shown  that  L  can  be  neither  greater 
than  nor  less  than  ep. 

First,  suppose  L  <  ep. 

Let  L  =  eK.     Then  K  <  p  (1) 

Now  inscribe  a  cylinder  the  perimeter  pr  of  whose 
right  section  is  greater  than  K.  This  is  possible  by  §  105. 

Then  ep'  >  eK.  (2) 

Hence,  ep'  is  greater  than  the  supposed  value  eK  of  L. 

But  this  is  impossible,  since  ep'  is  the  area  of  an  in- 
scribed prism  (§  107). 

Hence,  the  supposition  L  <  ep  leads  to  a  contradiction. 

Secondly,  the  supposition  that  L  >  ep  may  be  shown  to 
be  impossible  by  using  a  circumscribed  prism. 

Hence,  we  have  L  =  ep. 

109,  COROLLARY.   If  r  is  the  radius  of  any  circular 
cylinder,  and  e  is  an  element,  then  L  =  2  irre. 

In  the  case  of  a  right  circular  cylinder  e  =  h,  the  altitude, 
and  we  have  L  =  2 


52 


SOLID   GEOMETRY. 


110.  Definition.     Two  right  circular  cylinders  are  simi- 
lar if  they  are  generated  by  similar  rectangles  revolving 
about  corresponding  sides. 

111.  THEOREM.     The  lateral  areas  or  the  entire  areas 
of  two  similar  right  circular  cylinders  are  in  the  same 
ratio  as  the  squares  of  their  altitudes  or  as  the  squares 
of  the.  radii  of  their  bases. 


Suggestion.  If  h  and  h!  are  the  altitudes,  r  and  /  the 
radii,  L  and  L1  the  lateral  areas,  and  A  and  Af  the  total 
areas,  we  are  to  show  that 


_^ 
L'     A'      r'*     h'*' 

Make  use  of  the  following,  giving  each  in  detail. 
(1)  L  =  2  irrh,     (2)  A  =  2  irr(r  +  A), 


h 


r  4- 


112,  THEOREM.  The  volume  of  any  convex  cylinder 
is  equal  to  the  product  of  its  altitude  and  the  area  of  its 
base,  or  to  the  product  of  an  element  and  the  area  of  its 
right  section. 

Suggestions.  (1)  If  h  is  the  altitude  and  b  the  base, 
show  by  an  argument  similar  to  that  of  §  108,  that  F,  the 
volume,  can  be  neither  greater  nor  less  than  bh. 


PRISMS  AND  CYLINDERS.  53 

(2)  If  e  is  an  element  and  c  the  area  of  a  right  section, 
give  a  similar  argument  to  show  that  V  =  ec,  using  §  90. 
Give  all  the  steps  in  full. 

113,  COROLLARY.  If  a  cylinder  has  a  right  circu- 
lar section  whose  radius  is  r^  and  an  element  e,  then 
V=7rn2e. 

If  a  cylinder  has  a  circular  base  of  radius  r2,  and  an 
altitude  h,  then  V  = 


In  the  case  of  a  right  circular  cylinder,  r\  —  r2  and  h  =  e.  Hence, 
the  two  formulas  become  identical. 

114,  THEOREM.  The  volumes  of  two  similar  right 
circular  cylinders  are  in  the  same  ratio  as  the  cubes  of  the 
radii  of  their  bases  or  as  the  cubes  of  their  altitudes. 

Give  the  proof  in  full,  using  suggestions  similar  to  those 
given  in  §  111. 

PROBLEMS   AND   APPLICATIONS. 

1.  What  dimensions  of  a  cylinder  must  be  known  in  order  that  its 
lateral  area  may  be  computed?      State  fully  for  different  kinds  of 
cylinders. 

2.  What  dimensions  of  a  cylinder  must  be  known  in  order  that  its 
volume  may  be  computed  ? 

3.  If  the  base  of  a  cylinder  is  a  circle  with  radius  5  inches,  find 
its  volume  if  the  altitude  is  8  inches. 

4.  If  the  lateral  surface  of  a  cylinder  and  the  length  of  an  element 
are  known,  can  the  perimeter  of  a  right  section  be  found?      If  the 
lateral  area  is  400  TT,  and  an  element  15,  find  the  perimeter  of  a  right 
section. 

5.  The  diameter  of  a  right  circular  cylinder  is  8,  and  the  diagonal 
of  the  largest  rectangle  which  can  be  cut  from  it  is  16.      Find  its 
altitude. 

6.  The  volume  of  a  right  circular  cylinder  is  128  TT  cubic  inches 
and  its  altitude  is  equal  to  its  diameter.     Find  the  altitude  and  the 
diameter. 


54  SOLID   GEOMETRY. 

7.  If  the  diameter  of  a  right  circular  cylinder  is  equal  to  its  alti- 
tude, determine  the  diameter  so  that  the  total  area  of  the  cylinder  is 
equal  numerically  to  its  volume. 

8.  Find  the  diameter  of  a  right  circular  cylinder  if  the  total  area 
of  the  inscribed  regular  triangular  prism  is  equal  numerically  to  the 
volume  of  the  cylinder,  the  diameter  of  the  cylinder  being  equal  to 
its  altitude. 

9.  In  the  preceding  find  the  diameter  of  the  cylinder  if  the  volume 
of  the  prism  is  equal  numerically  to  the  total  area  of  the  cylinder. 

10.  Solve  a  problem  like  Ex.  8  if  a  regular  hexagonal  prism  is 
used  instead  of  a  triangular  prism. 

11.  Solve  a  problem  like  Ex.  9  if  a  regular  hexagonal  prism  is 
used  instead  of  a  triangular  prism. 

12.  A  rectangle  whose  sides  are  a  and  b  is  turned  about  the  side  a 
as  an  axis  and  then  about  tue  side  b.     Find  the  ratio  of  the  volumes 
of  the  two  cylinders  thus  developed. 

13.  Compare  the  total  surfaces  of  the  two  figures  developed  in 
Ex.  12. 

14.  Find  the  diameter  of  a  right  circular  cylinder  if  its  lateral 
area  is  equal  numerically  to  its  volume.     Does  the  result  depend 
upon  the  altitude  of  the  cylinder? 

15.  If  the  altitude  of  a  right   circular   cylinder   is  equal  to  its 
diameter,  find  the  ratio  of  the  numerical  values  of  its  total  area  and 
its  volume.     Does  this  depend  on  the  radius? 

16.  A  regular  octagonal  prism   is   inscribed %in   a  right  circular 
cylinder  whose  altitude  is  equal  to  the  diameter.     Find  the  difference 
between  the  volumes  of  the  cylinder  and  the  prism,  if  a  side  of  the 
octagon  is  4  inches. 

17.  A  cylindrical  tank  8  feet  in  diameter,  partly  filled  with  water, 
is  lying  on  its  side.     If  the  greatest  depth  of  the   water  is  6  feet, 
what  fraction  of  the  volume  of  the  tank  is 

filled  with  water? 

18.  In  the   preceding  problem  find  the 
fraction  of  the  volume  occupied  by  water  if 
the  width  of  the  top  of  the  water  along  a 
right  cross  section  of  the  tank  is  4  feet. 


PRISMS  AND   CYLINDERS. 


55 


THEOREMS  ON  PROJECTION. 

115,  The  projection  of  a  line-segment  on  a  given  line  was 
denned  in  §  57.     The  length  of  the  projection  will  now  be 
computed  in  terms  of  the  given  line-segment. 

116,  Definitions.    The   acute  angle  between  a  line-seg- 
ment and  a  given  line  on  which  it  is  projected  is  called 

the  projection  angle.  B 

L 


E.g.  ZAFC  or  its  equal  Z.BAE  found  by  drawing  A  E  II  CD. 

If  I  is  the  length  of  a  line-segment  and  p  the  length  of 

its  projection,  then  the  ratio  -^  is  called 

l 

the  cosine  of  the  projection  angle. 

E.g.  in  the  figure,  ^  =  cosine  Z  BA  E. 

JL- 7 £ 

117,    In  any  right  triangle  ABC,  either 

acute  angle,  as  ^  A,  is  the  projection  angle  between  the 
hypotenuse  and  the  side  adjacent  to  the  angle. 

Hence  the  cosine  of  an  acute  angle  of  a  right  triangle 
is  the  ratio  of  the  adjacent  side  to  the  hypotenuse. 

We  have  already  defined  the  sine  of  an  acute  angle  of  a 
right  triangle  as  the  ratio  of  the  opposite  side  to  the  hypote- 
nuse. See  §  279,  Plane  Geometry. 

We  now  define  the  tangent  of  an  acute  angle  of  a  right 
triangle  as  the  ratio  of  the  opposite  side  to  the  adjacent  side. 

Using  the  common  abbreviations,  sin,  cos,  and  tan,  we 
have  in  the  figure  : 


sin  A  =  -,  cos 
c 


=  -i  tan  A  =  - 
eo 


56  SOLID   GEOMETRY. 

118,  The  sine,  cosine,  and  tangent  are  of  great  impor- 
tance in  many  computations.     By  careful   measurement 
(and  in  other  ways)   their  values  may  be  computed  for 
any  acute  angle,  and  a  table  formed,  like  that  on  page  57. 

E.g.  if  in  the  figure  of  §  117  Z  A  =  35°  (measured  with  a  protractor), 
we  may  measure  AC,  AB,  and  BC,  and  thus  compute  the  ratios 

-,  -,  and  -,  and  find  the  values  of  sin  35°,  cos  35°.  tan  35°. 
c     c  b 

With  an  ordinary  ruler  it  will  not  usually  be  possible  to  make 
these  measurements  with  sufficient  accuracy  to  obtain  more  than  one 
decimal  place. 

Draw  an  angle  of  35°,  and  make  the  measurements  and  computa- 
tions, using  an  hypotenuse  of  various  lengths  (the  longer  the  better), 
and  show  that  the  results  do  not  depend  upon  the  length  of  hypote- 
nuse chosen. 

119.  EXERCISES. 

1.  Using  a  protractor,  construct  angles  of  10°,  30°,  50°,  70°,  and  by 
measurement  determine  the  sine,  cosine,  and  tangent  of  each. 

2.  Prove  that  the  cosine  of  any  given 
angle  is  the  same,  no  matter  what  point  is 
taken  in  either  side  from  which  to  let  fall 
the  perpendicular  to  the  other  side.     Prove 
the  same  for  the  tangent. 

3.  Show  that  if  the  hypotenuse  be  taken  one.  decimeter  in  length, 
then  the  length  of  the  side  adjacent,  measured  in  decimeters,  is  the 
cosine  of  the  angle,  and  the  length  of  the  side  opposite  is  the  sine  of 
the  angle. 

4.  Show  that   if  the   side   adjacent   be  taken  one   decimeter  in 
length,  the  length  of  the  side  opposite,  measured  in  decimeters,  is  the 
tangent  of  the  angle. 

5.  Without  any  direct  measurement,  show  how  to  compute  the 
three  ratios  for  each  of  the  angles,  30°,  45°,  60°. 

SUGGESTION.  Make  use  of  the  fact  that  if  one  acute  angle  in  a 
right  triangle  is  30°,  the  side  opposite  it  is  one  half  the  hypotenuse. 

6.  As  the  angle  is  made  smaller  and  smaller,  what  are  the  values 
approached  by  the  sine,  cosine,  and  tangent  ? 


PRISMS  AND   CYLINDERS. 


57 


7.  As  the  angle  is  made  more  and  more  nearly  90°,  what  are  the 
values  approached  by  the  sine  and  cosine?  Discuss  this  case  for  the 
tangent. 


TABLE  OF  SINES,  COSINES,  AND  TANGENTS. 


Ang. 

Sin 

Cos 

Tan 

Ang. 

Sin 

Cos 

Tan 

Ang. 

Sin 

Cos 

Tan 

0° 

0 

1 

0 

31° 

.515 

.857 

.601 

61° 

.875 

.485 

1.80 

1° 

.017 

1.000 

.017 

32° 

.530 

.848 

.625 

62° 

.883 

.469 

1.88 

2° 

.034 

.999 

.035 

33° 

.545 

.839 

.649 

63° 

.891 

.454 

1.96 

3° 

.052 

.999 

.052 

34° 

.559 

.829 

.675 

64° 

.899 

.438 

2.05 

4° 

.070 

.998 

.070 

35° 

.574 

.819 

.700 

65° 

.906 

.423 

2.14 

5° 

.087 

.996 

.087 

36° 

.588 

.809 

.727 

66° 

.914 

.407 

2.25 

6° 

.105 

.995 

.105 

37° 

.602 

.799 

.754 

67° 

.921 

.391 

2.36 

7° 

.122 

.993 

.123 

38° 

.616 

.788 

.781 

68° 

.927 

.375 

2.48 

8° 

.139 

.990 

.141 

39° 

.629 

.777 

.810 

69° 

.934 

.358 

2.61 

9° 

.156 

.988 

.158 

40° 

.643 

.766 

.839 

70° 

.940 

.342 

2:75 

10° 

.174 

.985 

.176 

41° 

.656 

.755 

.869 

71° 

.946 

.326 

2.90 

11° 

.191 

.982 

.194 

42° 

.669 

.743 

.900 

72° 

.951 

.309 

3.08 

12° 

.208 

.978 

.213 

43° 

.682 

.731 

.933 

73° 

.956 

.292 

3.27 

13° 

.225 

.974 

.231 

44° 

.695 

.719 

.966 

74° 

.961 

.276 

3.49 

14° 

.242 

.970 

.249 

45° 

.707 

.707 

1.00 

75° 

.966 

.259 

3.73 

15° 

.259 

.906 

.268 

46° 

.719 

.695 

1.04 

76° 

.970 

.242 

4.01 

16° 

.276 

.961 

.287 

47° 

.731 

.682 

1.07 

77° 

.974 

.225 

4.33 

17° 

.292 

.956 

.306 

48° 

.743 

.669 

1.11 

78° 

.978 

.208 

4.70 

18° 

.309 

.951 

.325 

49° 

.755 

.656 

1.15 

79° 

.982 

.191 

5.14 

19° 

.326 

.946 

.344 

50° 

.766 

.643 

1.19 

80° 

.985 

.174 

5.67 

20° 

.342 

.940 

.364 

51° 

.777 

.629 

1.23 

81° 

.988 

.156 

6.31 

21° 

.358 

.934 

.384 

52° 

.788 

.616 

1.28 

82° 

.990 

.139 

7.12 

22° 

.375 

.927 

.404 

53° 

.799 

.602 

1.33 

83° 

.993 

.122 

8.14 

23° 

.391 

.921 

.424 

54° 

.809 

.588 

1.38 

84° 

.995 

.105 

9.51 

24° 

.407 

.914 

.445 

55° 

.819 

.574 

1.43 

85° 

.996 

.087 

11.43 

25° 

.423 

.906 

.466 

56° 

.829 

.559 

1.48 

86° 

.998 

.070 

14.30 

26° 

.438 

.899 

.488 

57° 

.839 

.545 

1.54 

87° 

.999 

.052 

19.08 

27" 

.454 

.891 

.510 

58° 

.848 

.530 

1.60 

88° 

.999 

.035 

28.64 

28° 

.469 

.883 

.532 

59° 

.857 

.515 

1.66 

89° 

1.000 

.017 

57.29 

29° 

.485 

.875 

.554 

60° 

.866 

.500 

1.73 

90° 

1.000 

0 

30° 

.500 

.866 

.577 

58  SOLID   GEOMETRY. 

120,  THEOREM.     The  length  of  the  projection  of  a 
line-segment  upon  a  given  line  is  equal  to  the  length  of 
the  line-segment  multiplied  ~by  the  cosine  of  the  projec- 
tion angle. 

Given  the  projection  p  of  the 
line-segment  /  on  the  line  CD, 
with  the  projection  angle  A. 

To  prove  that  p  =  I  cos  A.  Af^~ — J™~ E 

Proof :    By   definition    we 

have  y  =  cos  A.  c  ~D 

Hence,  p  =  I  cos  A. 

121,  EXERCISES. 

1.  Find  the  cosines  of  the  angles  35°  30',  54°  15',  15°  45'. 
SUGGESTION.     The  cosine  of  35°  30'  lies  between  cos  35°  and  cos  36°. 

We  assume  that  it  lies  halfway  between  these  numbers.  This  as- 
sumption, while  not  quite  correct,  is  very  nearly  so  for  small  differences 
of  angles,  as  in  this  case,  where  the  total  difference  is  only  one  degree. 
From  the  table  cos  35°  =  .819,  cos  36°  =  .809. 

The  number  midway  between  these  is  .814,  which  we  take  as  the 
cosine  of  35°  30'. 

This  process  is  called  interpolation.  A  similar  process  is  used  for 
sines  and  tangents. 

2.  Find  the  tangents  of  the  angles  25°  20',  47°  45',  63°  40'. 

3.  Find  the  angle  whose  tangent  is  1.74. 

SOLUTION.  From  the  table  we  have  tan  60°  =  1.73  and  tan  61°  =  1.80. 
Hence,  the  required  angle  must  lie  between  60°  and  61°.  Moreover, 
the  number  1.74  is  one  seventh  the  way  from  1.73  to  1.80.  Hence,  we 
assume  the  angle  to  lie  one  seventh  the  way  from  60°  to  61°,  which 
gives  60°  +  }  x  1°  =  60° +  9'  nearly.  The  required  angle  is  60°  9'. 

4.  Find  the  angles  whose  sines  are  .276 ;  .674 ;  .437. 

5.  Find  the  angles  whose  cosines  are  .940 ;  .094 ;  .435. 


PRISMS  AND   CYLINDERS. 


59 


179  ft 


6.  Find  the  angles  whose  tangents  are  .781 ;  1.41 ;  3.64. 

Notice  that  as  an  angle  increases,  its  sine  and  tangent 
both  increase,  but  its  cosine  decreases. 

7.  At  what  angle  with  the  horizontal  must 
the  base  of  a  right  circular  cylinder  be  tilted  to 
make  it  just  topple  over  if  its  diameter  is  6  feet 
and  its  altitude  8  feet? 

SUGGESTION.  The  center  of  gravity  is  at  the 
middle  point  C  of  the  axis  of  the  cylinder.  The 
base  must  be  tilted  so  that  the  line  A  C  becomes 
vertical. 

8.  The  Leaning  Tower  of  Pisa  is  179  feet  high 
and  31  feet  in  diameter.     It  now  leans  so  that  a 
plumb  line  from  the  top  on  the  lower  side  reaches 
the  ground  14  feet  from  the  base. 

At  what  angle  is  its  side  now  inclined  from 
the  vertical  ?  At  what  angle  would  its  side  have 
to  incline  from  the  vertical  before  it  would  topple 
over  ?  v — _— ^14  "it. 

9.  A  four-inch  hole  is  cut  in  a  board,  and  a  ball  8  in.  in  diameter 
is  made  to  rest  on  it.     At  what  angle 

must  the  board  be  held  so  that  the  ball 
will  just  roll  out  of  the  hole  ? 

SUGGESTION.  The  board  must  be 
held  so  that  the  line  OA  becomes  ver- 
tical ;  that  is,  the  board  must  be  tipped 
at  an  angle  equal  to  /.  BOA. 

10.  Using  a  ball  8  inches  in  diameter,  what  must  be  the  radius  of 
the  hole  in  the  board  of  the  preceding  problem  so  that  the  ball  shall 
just  roll  out  when  the   board  is  inclined 

at  an  angle  of  45°  to  the  horizontal? 

11.  If  the  figure  ABCD-H  is  a  cube, 
find  each  of  the  following  angles:  ^ECA, 
ZAEC. 

Check  the  results  found  by  using  the 
fact  that  the  sum  of  the  angles  of  a  tri- 
angle is  180°. 


60  SOLID   GEOMETRY. 

122,  THEOREM.  The  altitude  of  an  oblique  prism  or 
cylinder  is  equal  to  an  element  multiplied  by  the  cosine 
of  the  angle  between  the  plane  of  the  base  and  that  of  a 
right  section. 


G 


E         -^ 


Given  the  oblique  cylinder  with  base  b  and  right  section  c,  and 
let  BE  be  a  perpendicular  between  the  bases. 

Consider  the  plane  determined  by  BE  and  the  element 
AB. 

This  plane  is  J_  to  the  plane  of  b  and  also  to  the  plane 
ofc.  (Why?) 

Hence,  it  is  _L  to  the  line  of  intersection  of  the  planes  of 
b  and  c.  (Why?) 

Let  this  plane  cut  the  planes  of  b  and  c  in  GD  and  FD 
respectively,  D  being  the  point  in  their  line  of  intersection. 

Then  Z  D  is  the  measure  of  the  dihedral  angle  between 
the  planes  of  b  and  c.  (Why?) 

To  prove  that  BE  =  AB  -  cos  D. 

Proof  :    We  have  BE  =  AB  -  cos  Z  ABE.       (Why?) 

But  Z.D  =  ^.ABE.  (Why?) 

Hence,  BE  —  AB  •  cos  D. 

The  argument  is  similar  for  any  oblique  prism, 

123,  COROLLARY.  The  dihedral  angle  betioeen  the 
planes  of  the  base  and  a  right  section  of  an  oblique 
cylinder  or  prism  is  equal  to  the  angle  between  an 
element  and  the  altitude. 


PRISMS  AND   CYLINDERS. 


61 


124, 


EXERCISES. 


n 


1.  Given  a  line-segment  10  inches  long.     Find  the  length  of  its 
projection  on  a  plane  if  the  projection  angle  is  20°.     If  the  angle  is 
30°,  45°,  60°,  90°,  0°. 

2.  A  kite  string  forms  an  angle  of  40°  with  the  ground.     The 
distance  from  the  end  of  the  string  to  a  point  directly  beneath  the 
kite  is  200  ft.     Find  the  length  of  the  string  and  the  perpendicular 
height  of  the  kite. 

3.  The  altitude  of  an  oblique  prism  is  15  inches.     Find  the  length 
of  an  element  if  it  makes  an  angle  of  45°  with  the  perpendicular 
between  the  bases. 

4.  A  right  section  of  a  cylinder  makes  an  angle  of  20°  with  the 
plane  of  the  base.    Find  the  ratio  between  the  altitude  and  an  element. 

5.  Show  that  the  theorem  of  §  122  holds  for  the  special  case  of  a 
right  prism  or  cylinder. 

6.  Prove    that    by    joining    the 
middle  points  of  six  edges  of  a  cube, 
as    shown   in    the  figure,   a  regular 
hexagon  is  formed. 

7.  Prove  that  in  the  preceding 
example   the   plane   of   the   regular 
hexagon,  KLMNOP,  is  perpendicu- 
lar  to  the  diagonal  DF. 

8.  How  large  a  cube  will  be  re' 
quired  from  which  to  cut  a  stopper 
for  a  hexagonal  spout,  each  of  whose 
sides  is  4  inches  ? 

9.  In  the  figure  find  the  angle  KQH. 

SUGGESTION.     Let  a  be  a  side  of  the  cube.     Compute  KH,  KQ,  and 
HQ  in  terms  a.     Note  that  Z  QKH  =  rt.  Z. 

10.  Find  the  area  of  the  projection  of  the  hexagon  KLMNOP  on 
the  face  BCGF.     Note  that  this  projection  equals  the  whole  square 
less  A  NCO  +  A  KEL.    See  §  125. 

11.  Find  the  area  of  the  hexagon  in  terms  of  the  side  a  of  the  cube. 

12.  By  means  of  the  theorem  of  §  126  and  the  results  in  Exs.  10 
and  11  find  the  dihedral  angle  formed  by  the  planes  BCG  and  MOK. 


62 


SOLID  GEOMETRY. 


PROJECTION  OF  A  PLANE-SEGMENT. 

125.  Definition.     If  from  each  point  in  the  boundary  of 
a  plane-segment  a  perpendicular  is  drawn  to  a  given  plane, 
the  locus  of  the  feet  of  these  perpendiculars  Avill  bound  a 
portion  of  the  plane,  which  is  called  the  projection  of  the 
plane-segment  on  the  given  plane. 

E.g.  the  plane-segment  A'B'C'  in  the 
plane  N  is  the  projection  of  the  plane- 
segment  ABC  from  the  plane  M  upon  N. 

The  angle  of  projection  is  the  angle 
between  the  pjanes  M  and  N. 

126,  THEOKEM.      The  area  of  the  projection  of  a 
plane-segment  on  a  plane  is  equal  to  the  area  of  the 
plane-segment  multiplied  by  the  cosine  of  the  projection 
angle. 


Proof:  Let  the  boundary  of  the  given  plane-segment  b 
be  any  convex  polygon  or  closed  curve. 

Using  this  polygon  or  curve  as  a  directrix  and  a  line  per- 
pendicular to  the  given  plane  as  a  generator,  develop  a 
prismatic  or  cylindrical  surface.  The  given  plane  will  cut 
this  surface  in  a  right  section  whose  area  we  denote  by  c. 

Now  cut  the  surface  by  a  plane  parallel  to  6,  forming  the 
upper  base  bf  of  a  prism  or  cylinder  whose  altitude  is  h, 
edge  g,  and  volume  V. 


PRISMS  AND  CYLINDERS. 


63 


Then  c  is  the  projection  of  b  upon  the  given  plane,  and 
Z  1  =  Z  2  is  the  projection  angle. 
We  are  to  show  that  c  —  b  cos  Z  1. 
We  know  that  V  =  ce  =  bh.  (Why  ?) 

But  h  =  e  cos  Z  2. 

Hence,  ce  =  be  cos  Z  2. 

That  is,  c  =  b  cos  Z  2  =  b  cos  Z  1. 

NOTE.  The  foregoing  theorem  may  be  proved  directly  in  case  the 
plane-segment  is  a  rectangle  with  one  side1 
parallel  to  the  line  of  intersection  of  the  two 
planes.  In  the  figure  let  S  be  the  given  rec- 
tangle and  S'  its  projection,  with  AB  II  to  the 
line  of  intersection  of  the  planes  in  which  S 
and  S'  lie,  and  /.  1  the  angle  between  them. 


E 


Then 
and 

But 
and 

But 


S  =  AB • EC 

S'  =  A'B' .  B'C'. 
AB  =  A'B' 

BE  =  B'C'.  (Why?) 

BE  =  BC  •  cos  Z 1 

(Why?) 


and  S1  =  A'B'  •  B'C' =  AB  •  BC  cos  ^L 

That  is,  S1  =  S  cos  Z  1. 

In  the  case  of  any  plane-segment,  rectangles  may  be  inscribed  in  it 
in  this  position  and  their  number  increased  indefinitely,  so  that  their 
sum  will  approach  more  and  more  nearly  to  the  area  of  the  plane- 
segment,  and  in  this  way  it  may  be  shown  to  any  desired  degree  of 
approximation  that  the  projection  of  a  plane-segment  equals  the  given 
plane-segment  multiplied  by  the  cosine  of  the  projection  angle. 

127,  An  important  special  case  of  the  theorem  of  §  126 
is  the  area  of  the  figure  obtained  by  projecting  a  circle 
upon  a  plane  not  parallel  to  the  plane  of  the  circle. 

Definition:  The  figure  obtained  by  projecting  a  circle 
upon  a  plane  not  parallel  to  the  plane  of  the  circle,  nor  at 
right  angles  to  it  is  called  an  ellipse. 


64 


SOLID   GEOMETRY. 


128,  Area  of  the  Ellipse.  —  In  the  figure  two  planes,  M 
and  Jif',  meet  in  a  line  PQ.  The  circle  O  in  M  has  a  diameter 
AB  II  PQ  and  a  diameter  CD  _L  PQ. 

In  projecting  the  whole  figure  upon  the  plane  M1  the 
diameter  AB  projects  into  its  equal  A'B',  while  CD  projects 
into  c'Df  so  that  C'D'=  CD  cos  Z  1. 


By  theorem  §  126  the  area  of  the  ellipse  equals  the  area 
of  the  circle  multiplied  by  cos  ^  1. 

Hence,  Trr2  •  cos  Z  1  is  the  area  of  the  ellipse. 


But  r  cos 


=  o'cf  and  r  =.  o'B1. 


(§120) 


Hence,  the  area  of  the  ellipse  is  TT  •  o'c'  X  O'B'  . 

The  segments  A'B'  and  C'D'  are  called  respectively  the 
major  and  minor  axes  of  the  ellipse,  and  o'B1  and  o'c'  the 
semimajor  and  the  semiminor  axes.  These  latter  are  usu- 
ally denoted  by  a  and  b. 

Hence,  the  area  of  the  ellipse  is  irab. 

Note  that  when  a  and  b  are  equal,  the  ellipse  becomes  a  circle,  and 
this  formula  reduces  to  ira?  as  it  should. 

It  may  be  of  interest  to  note  that  the  problem  of  finding  the  length 
of  the  ellipse  is  very  much  more  difficult,  and  can  be  solved  only  by 
means  of  higher  mathematics. 


PEISMS  AND   CYLINDERS.  65 

129,  The  figure  of  §  128  may  also  be  regarded  as  repre- 
senting a  cylindrical  surface  of  which  the  ellipse  with 
center  o'  is  a  right  section. 

It  is  also  true  that  if  we  start  with  a  circular  cylinder, 
that  is,  one  whose  right  section  c  is  a  circle,  then  every 
oblique  section  of  it,  as  d,  is  an  ellipse. 

The  minor  axis  of  such  an 
ellipse  will  be  the  diameter 
2r  of  the  circle,  and  the 
major  axis  2  r  -5-  cos  Z  1. 

Thus,  if  a  circular  cylinder  of 
diameter  6  in.  is  cut  by  a  plane 
making  an  angle  of  30°  with  the 
right  section  c,  then  the  section  c' 
thus  made  is  an  ellipse  whose  axes 
are  6  in.  and  6  -j-  cos  30°  =  6  -=-  .866  =  6.93  in.  nearly. 

Hence,  the  area  of  this  elliptical  section  is 

-nab  =  3.14  x  6.93  x  6  =  130.56  sq.  in. 

SUMMARY  OF   CHAPTER  H. 

1.  Make  a  list  of  the  definitions  on  prisms  and  also  a  list  of  those 
on  cylinders  and  compare  them. 

2.  Make  a  list  of  the  theorems  on  prisms  and  also  of  those  on 
cylinders  and  compare  them. 

3.  State  the  axioms  of  this  chapter  and  note  that  they  all  refer 
to  areas  and  volumes  which  involve  curved  surfaces.     Compare  these 
with  the  axioms  on  the  circle  in  Plane  Geometry. 

4.  Make   a  list  of  all  the  formulas  given  by  the   theorems   and 
corollaries  of  this  chapter. 

5.  Make  an  outline  of  the  definitions  and  theorems  concerning  pro- 
jections of  lines  and  surfaces. 

6.  Explain  how  the  area  of  an  ellipse  is  obtained  either  by  project- 
ing it  into  a  circle  or  by  projecting  a  circle  into  it. 

7.  Make  a  list  of  the  applications  in   this  chapter  which  have 
appealed  to  you  as  interesting  or  practical  or  both.     Return  to  this 
question,  after  studying  the  problems  and  applications  which  follow. 


66  SOLID  GEOMETRY. 

• 
PROBLEMS  AND  APPLICATIONS. 

1.  Given  a  right  circular  cylinder  the   radius   of   whose  base   is 
6  inches.     Find  the  area  of  an  oblique  cross  section  inclined  at  an 
angle  of  45°  to  the  plane  of  the  base. 

2.  Given  an  oblique  circular  cylinder  the  radius  of  whose  right 
section  is  10  inches.     Find  the  area  of  the  base  if  it  is  inclined  at  an 
angle  of  60°  to  the  right  section. 

3.  If  an  oblique  circular  cylinder  has  an  altitude  h,  an  element  e, 
radius  of  right  section  r,  and  Z  A  the  inclination  of  the  base  to  the 
right  section,  express  the  volume  in  two  ways  and  show  that  these  are 
equivalent. 

4.  A  six-inch  stovepipe  has  a  45°  elbow,  that  is, 

it  turns  at  right  angles.     (The  angle  CAB  is  called 


the  elbow  angle.)     Find  the  area  of  the  cross  section  B 

at  AB.     Likewise  if  it  has  a  60°  elbow  angle. 

5.  At  what  angle  must  the  damper  in  a  circular 
stovepipe  be  turned  in  order  to  obstruct  just  half  the 
right  cross  sectional  area  of  the  pipe? 

SUGGESTION.  The  damper  must  be  turned  through  an  angle  such 
that  the  area  of  the  projection  of  the  damper  upon  a  right  cross  section 
is  equal  to  half  that  cross  section. 

6.  The  comparatively  low  temperature  of  the  earth's  surface  near 
the  pole,  even  in  summer,  when  the  sun  does  not  set  for  months,  is  due 
largely  to  the  obliqueness  with  which  the  sun's  rays  strike  the  earth. 
That  is,  a  given  amount  of  sunlight  is  spread  over  a  larger  area  than 
in  lower  latitudes. 

Thus,  if  in  the  figure  D'C  is  a  horizontal  line,  and  D'D  the  direc- 
tion of  the  sun's  rays,  then  a  beam  of  light  whose  right  cross  section 
is  A  BCD  is  spread  over  the  rectangle 
A' BCD'.  In  other  words,  a  patch 
of  ground  A' BCD'  receives  only  as 
much  sunlight  as  a  patch  the  size  of 
A  BCD  receives  when  the  sun's  rays 
strike  it  vertically. 

Area  A  B  CD  =  area  A'B  CD'  cos  Z 1, 
or 

=  are&ABCD  x 


cos  Zl 


PMISMS  AND  CYLINDERS.  67 

Hence,  each  unit  of  area  in  A'  BCD'  receives  cos  Z  1  times  as  much 
light  as  a  unit  in  A  BCD. 

Hence,  to  compare  the  heat-producing  powers  of  sunlight  in  any 
latitude  with  that  at  the  place  where  the  sun's  rays  fall  vertically,  we 
need  to  know  how  the  projection  angle,  Z  1,  is  related  to  the  difference 
in  latitude  of  the  two  places. 

7.  If  Z  1  =  30°,  compare  the  amount  of  heat  received  by  a  unit  of 
area  in  A  B  CD  and  A'  BCD'. 

8.  What  must  Z  1  be  in  order  that  a  unit  of  area  in  A'  BCD' 
shall  receive  only  ^  as  much  light  as  a  unit  in  A  BCD'! 

9.  The  figure  represents  a  cross  section  of  the  earth  with  an  indi- 
cation of  the  direction  of  the  rays  of  light  as  they  strike  it  at  the 
summer  solstice  when  they  are  vertical  at  A,  the  tropic  of  Cancer.     B 
represents  the  latitude  of  Chicago,  Cthe  polar  circle,  and  P  the  north 
pole.      The    angles    PDE, 

CGF,  BKH  represent  the 
projection  angle,  Z  1,  for 
the  various  latitudes. 

Prove     that    Z  PDE  = 
Z.POK, 


That  is,  Z  1  for  each  place 
is  the  latitude  of  that  place 
minus  the  latitude  of  the 
place  where  the  sun's  rays  are  vertical. 

10.  Find  the  relative   amount  of  sunlight  received  by  a  unit  of 
area  at  the  tropic  of  Cancer  and  at  the  north  pole  at  the  time  of  the 
summer  solstice. 

SUGGESTION.     The  required  ratio  is  -  =  -  =  -  . 

cos  Z  1      cos  .66£°      .399 

11.  Find  the  ratio  between  the  amount  of  light  received  by  a  unit 
of  the  earth's  area  at   Chicago  and  at  the  tropic  of  Cancer  at  the 
time  of  the  summer  solstice. 

12.  Find   the   same  ratio  for  the  polar  circle  and  the  tropic   of 
Cancer   at   the  spring  equinox,  when   the  sun  is  vertical   over   the 
equator. 

13.  Find  the  same  ratio  for  the  equator  and  Chicago  at  the  winter 
solstice  when  the  sun  is  vertical  at  latitude  23£°  south. 


CHAPTER   III. 


PYRAMIDS  AND  CONES. 

130,  Definitions.     Given  a  convex  polygon  and  a  fixed 
point  not  in  its  plane.     If  a  line  through  the 

fixed  point  moves  so  as  always  to  touch  the 
polygon  and  is  made  to  traverse  it  completely, 
the  line  is  said  to  generate  a  convex  pyramidal 
surface. 

The  moving  line  is  the  generator  of  the  sur- 
face, and  in  any  of  its  positions  it  is  an  element 
of  the  surface.  The  guiding  polygon  is  the 
directrix,  and  the  fixed  point  the  vertex.  A 
pyramidal  surface  has  two  parts,  called  nappes,  on  op- 
posite sides  of  the  fixed  point.  Compare  definitions  in 
§148. 

A  polyhedral  angle  is  a  pyramidal  surface  of 
one  nappe.     See  §  63. 

131,  That  part  of  a  pyramidal  surface 
included  between  the  fixed  point  and  a 
plane  cutting  all  its  elements,  together 
with   the   intercepted   segment   of   the 
plane,  is  called  a  pyramid. 

The  intercepted  plane-segment  is  the  base  of  the  pyra- 
mid, and  the  part  of  the  pyramidal  surface  between  the 
base  and  the  vertex  is  the  lateral  surface. 

68 


PYRAMIDS  AND   CONES. 

The  lateral  surface  is  composed  of  triangles 
having  a  common  vertex  at  the  vertex  of 
the  pyramid,  and  having  as  bases  the  sides 
of  the  polygonal  base.  The  sides  common 
to  two  such  triangles  are  the  edges  of  the 
pyramid. 

Pyramids  are  classified  according  to  the  shape  of  the 
base,  as  triangular,  quadrangular,  pentagonal,  etc. 

A  pyramid  having  a  triangular  base  has,  in  all,  four 
faces,  and  is  called  a  tetrahedron.  In  this  case  every  face 
is  a  triangle,  and  any  one  may  be  taken  as  the  base. 

The  altitude  of  a  pyramid  is  the  perpendicular  distance 
from  the  vertex  to  the  base. 

A  regular  right  pyramid,  or  simply  a  regular  pyramid,  is 
one  whose  base  is  a  regular  polygon  such  that  the  perpen- 
dicular from  the  vertex  upon  it  meets  it  at  the  center. 

EXERCISE.  Show  that  the  faces  of  a  regular  right  pyramid  are 
congruent  isosceles  triangles,  and  hence  have  equal  altitudes. 

The  slant  height  of  a  regular  right  pyramid  is  the  alti- 
tude of  any  one  of  its  triangular  faces. 

132,  THEOREM.  The  lateral  area  of 
a  regular  right  pyramid  is  equal  to 
one  half  the  product  of  its  slant  height 
and  the  perimeter  of  the  base. 

Suggestion.  Calling  L  the  lateral  area,  s  =  KP  the  slant 
height,  and  p  =  AB  +  BC  +  CD  +  DE  +  EA,  the  perimeter, 
show  that 


70 


SOLID   GEOMETRY. 


133,    THEOREM.     If  a  pyramid  is   cut   ~by  a  plane 
parallel  to  the  base : 

(1)  The  edges  and  the  altitude  are  divided  in  the 
same  ratio. 

(2)  The  polygonal  section  is  similar  to  the  base. 

(3)  The  areas  of  this  section  and  of  the  base  are  in 
the  same,  ratio  as  the  squares  of  their  perpendicular 
distances  from  the  vertex. 


Outline  of  proof  :  Given  ABODE  II  A'B'C'D'E'. 


(1)    To  prove  that 


"piift          ft  A^          "PTl^ 

-  -  =  -  -  =  -  ,  etc.,  pass  another 
PM        PA        PB 

plane  through  P  parallel  to  the  base  and  then  use  §  37. 
(2)  To  prove  that  ABODE  ~  A'B'C'D'E',  we  show  that 


B  =  Z  Br,  etc.,  and  also 


,  etc. 


AB        BC 
(3)  Calling  the  area  of  the  cross  section  br  and  that  of 


the  base  0,  we  are  to  prove  that  — 

0 


PM' 
~PM' 


,  and  for  this 


we  need  to  show  that 


AB 


PA' 
PA 


PM' 

PM 


Give  all  the  steps  in  detail. 


PYRAMIDS  AND   CONES.  71 

134,  COROLLARY.  If  two  pyramids  have  equal  alti- 
tudes and  bases  of  equal  areas  lying  in  the  same  plane, 
the  sections  made  ~by  a  plane  parallel  to  the  plane  of 
the  liases  have  equal  areas. 

Suggestion.  If  t  and  t'  are  the  areas  of  the  sections  and 
b  and  bf  those  of  the  bases,  show  by  the  theorem  that 

-  =  4»  and  hence  that  t  =  *'  if  b  =  b'. 
b      b 

135.  EXERCISES. 

1.  What  is  the  slant  height  of  a  regular  right  pyramid  if  its 
lateral  area  is  160  sq.  in.  and  the  perimeter  of  its  base  is  20  in.? 

2.  What  is  the  perimeter  of  the  base  of  a  regular  right  pyramid 
if  its  lateral  area  is  250  sq.  in.  and  its  slant  height  is  17  in.  ? 

3.  How  could  you  find  the  lateral  area  of  a  regular  right  pyramid  ? 
Of  any  irregular  pyramid  ?    What  measurements  would  be  necessary 
in  each  case  ?     Why  are  fewer  measurements  needed  in  the  case  of  a 
regular  right  pyramid? 

4.  The  base  of  a  regular  right  pyramid  is  a  regular   hexagon 
whose  side  is  8  ft.     Find  the  lateral  area  if  the  altitude  of  the  pyra- 
mid is  6  ft. 

5.  The  lateral  area  of  a  regular  right  hexagonal  pyramid  is  48 
sq.  ft.  and  the  slant  height  is  12  ft.    Find  the  altitude  of  the  pyramid. 

6.  The  base  of  a  regular  right  pyramid  is  a  square  whose  side  is 
16  ft.,  and  the  altitude  of  the  pyramid  is  6  ft.     Find  the  lateral  area. 

7.  A  pyramid  with  altitude  8  and  a  base  whose  area  is  36  is  cut  by 
a  plane  parallel  to  the  base  so  that  the  area  of  the  section  is  18  sq.  in. 
Find  the  distance  from  the  base  to  the  cutting  plane. 

8.  If  the  altitude  of  a  pyramid  is  A,  how  far  from  the  base  must  a 
plane   parallel  to  it  be  drawn  so  that  the  area  of  its  cross  section 
shall  be  half  that  of  the  base  of  the  pyramid  ? 

9.  In  a  regular  right  pyramid  a  plane  parallel  to  the  base  cuts  it 
so  as  to  make  a  section  whose  area  is  one  half  that  of  the  base.    Find 
the  ratio  between  the  lateral  area  of  the  pyramid  and  that  of  the  small 
pyramid  cut  oif  by  the  plane. 


SOLID  GEOMETRY. 


136,  Definition.     A  triangular  pyramid  is  cut  by  a  series 
of  planes  parallel  to  the  base,  including  one  through  the 
vertex  and  also  the 

one  in  which  the  base 
lies. 

Through  the  lines 
of  intersection  of 
these  planes  with  one 
of  the  faces,  planes 
are  constructed  par- 
allel  to  the  opposite 
edge,  thus  forming 
a  set  of  prisms  all  lying  within  the  pyramid,  as  a1  ',  £>',  c'  , 
in  pyramid  p',  or  a  set  lying  partly  outside  the  pyramid, 
as  #,  5,  <?,  d  in  pyramid  P. 

The,  inner  prisms  thus  constructed  are  called  a  set  of 
inscribed  prisms,  and  the  other  a  set  of  circumscribed 
prisms. 

137,  Axiom  VII.     A  pyramid  has  a  definite  volume 
which  is  less  than  the  combined  volume  of  any  set  of 
circumscribed  prisms  and  greater  than    that   of  any 
set   of  inscribed  prisms. 


138, 


EXERCISES. 


1.  In  the  figure  above  prove  that  prisms  d  and  c'  are  equal  in 
volume.     Also  that  c  =  b't  etc. 

2.  If  the  area  of  the  base  of  pyramid  P  is  12  sq.  in.,  what  is  the 
altitude  of  prism  a  if  its  volume  is  1  cu.  in.?    If  its  volume  is  ^  cu.  in.? 

3.  If  the  altitude  of  the  pyramid  in  the  preceding  exercise  is  16  in., 
into  at  least  how  many  equal  parts  must  it  be  divided  if  the  volume 
of  prism  a  is  to  be  less  than  1  cu.  in.  ?    less  than  .01  cu.  in.  ?     Is  it 
possible  to  divide  the  altitude  of  the  pyramid  into  a  sufficiently  large 
number  of  equal  parts  to  make  the  volume  of  prism  a  as  small  as  we 
like? 


PYRAMIDS  AND   CONES. 


73 


139,  THEOREM.  If  two  triangular  pyramids  have 
equal  altitudes  and  bases  of  equal  areas,  their  volumes 
are  equal. 


P  and  f,  in  which  PM=P'Mt,  and  the  bases 
ABC  and  A'B'C'  have  equal  areas. 

To  prove  that  P  and  pr  have  equal  volumes. 

Proof :  Divide  PM  and  P'M'  into  the  same  number  of 
equal  parts,  and  using  these  division  points,  construct  a 
set  of  circumscribed  prisms  for  P  and  a  set  of  inscribed 
prisms  for  Pf. 

Then  a'  =  5,  b'  =  c,  c'  =  d.  (See  §  94.) 

Denote     a  +  b  +  c+  d 'by  F  and  a'  +  6'  +  c'  by  Vr. 
Then  F-F'  =  a.      (Why?)  (1) 

If  P  differs  at  all  in  volume  from  p',  let  P  be  the  greater, 
and  let  the  difference  be  some  fixed  number,  IT,  so  that 

P  -  P'  =  K.  (2) 

But  from  (1)  P  —  Pf  <  a,  since  P  <  F  and  Pf  >  V1 . 

Now  a  can  be  made  less  than  K  by  taking  the  divisions 
on  PM  small  enough. 

Hence,  P  -  P'  <  K.  (3) 

Thus  (3)  contradicts  (2),  and  hence  the  supposition 
that  P  and  P;  differ  in  volume  is  impossible. 


74 


SOLID   GEOMETRY. 


140,    THEOREM.     The  volume  of  a  triangular  pyramid 
is  one  third  of  the  product  of  its  base  and  altitude. 


Given  the  triangular  pyramid  E-ABC.  Let  h,  b,  and  V  be  the 
numerical  measures  respectively  of  the  altitude  EM,  the  base  ABC 
and  the  volume. 

To  prove  that  v  =  ^bh. 

Proof :  Construct  on  the  base  ABC  a  triangular  prism 
with  altitude  h  and  lateral  edge  EB. 

This  prism  may  be  cut  into  three  pyramids,  as  shown 
in  the  figure  to  the  right,  by  the  plane  sections  through 
DEC  and  AEG.  See  Note,  §  68. 

The  pyramids  E-ABC  and  C-DEF  have  the  same  volume 
(§  139). 

Likewise  the  pyramids  E-ACD  and  E-CFD  have  the  same 
volume  (§  139). 

But  C-DEF  and  E-CFD  are  only  different  notations  for 
the  same  pyramid. 

Hence,  E-ABC  =  C-DEF  =  E-A  CD. 

That  is,  E-ABC  is  one  third  of  the  prism. 

But  the  volume  of  prism  =  bh.  (Why?) 

Hence,  F=  volume  of  pyramid  =  j  bh. 

State  in  detail  the  reasons  for  each  step. 


PYRAMIDS  AND   CONES. 


75 


141,   THEOREM.     The  volume  of  any  pyramid  is  one 
third  of  the  product  of  its  base  and  altitude. 


Given  the  pyramid  P-ABCDE.  Let  7,  6,  and  h  be  the  numeri- 
cal measures  respectively  of  the  volume,  base,  and  altitude. 

To  prove  that  v  =  |  bh. 

Proof :  By  means  of  the  diagonal  planes  PAC  and  PAD, 
divide  the  given  pyramid  into  three  triangular  pyramids. 

Complete  the  proof. 

142,  COROLLARIES.     1.    The   volumes    of  any    two 
pyramids  having  the  same  or  equal  altitudes  are  in 
the  same  ratio  as  the  areas  of  their  bases. 

2.  The  volumes  of  any  two  pyramids  having  the 
same  or  equal  bases  are  in  the  same  ratio  as  their 
altitudes. 

143,  EXERCISES. 

1.  The  altitude  of  a  certain  pyramid  is  14  in.  and  its  volume  is 
380  cu.  in.     Find  the  area  of  its  base. 

2.  The  area  of  the  base  of  a  pyramid  is  48  sq.  ft.  and  its  volume 
260  cu.  ft.     Find  its  altitude. 

3.  Find  the  locus  of  the  vertices  of  pyramids  having  the  same 
base  and  equal  volumes. 

4.  A  diagonal  of  the  square  base  of  a  regular  right  pyramid  is 
7\/2  in.  and  its  volume  147  cu.  in.     Find  its  altitude  and  lateral  area. 


76  SOLID   GEOMETRY. 

5.  A  flower  bed  is  in  the  form  of  a  regular  right  pyramid,  with  a 
square  base  5  ft.  on  a  side.     The  altitude  is  2  ft.     Find  the  number 
of  cubic  feet  of  soil  in  its  construction. 

6.  A  tent  is  to  be  made  in  the  form  of  a  right  pyramid,  with  a 
regular  hexagonal  base.     If  the  altitude  is  fixed  at  15  ft.,  what  must 
be  the  side  of  the  base  in  order  that  the  tent  may  inclose  350  cu.  ft. 
of  space  ? 

7.  Two   marble  ornaments  of  equal  altitudes  are  pyramidal  in 
form.     One  has  a  square  base  2  in.  on  a  side  and  the  other  a  regular 
hexagonal  base  1  in.  on  a  side.     Compare  their  volumes. 

8.  Two  monuments  having  bases  of  equal  areas  are  pyramidal  in 
shape,  one  being  15  ft.  high  and  the  other   18  ft.     Compare   their 
volumes. 

9.  If  the  base  and  the  volume  of  a  pyramid  are  known,  is  it  possi- 
ble to  determine  its  lateral  area  ? 

10.  Given  a  pyramid  with  rectangular  base.     By  how  much  is  its 
volume  multiplied  if  the  length  and  width  of  the  base  and  also  the 
altitude  are  each  multiplied  by  2 ;  by  3;  by  any  number  n  ? 

11.  Given  a  pyramid  with  altitude  10  and  a  regular  hexagonal 
base,  each  of  whose  sides  is  5.     By  how  much  is  its  volume  multiplied 
if  each  side  of  the  base  and  also  the  altitude  is  multiplied  by  2  ;  by  3 ; 
by  4  ;  by  any  number  n  ? 

For  a  general  statement  of  the  law  exemplified  in  these  exercises, 
see  §  195. 

144,  Definitions.  The  figure  formed  by 
the  base  of  a  pyramid,  any  cross  section,  and 
the  portion  of  the  lateral  faces  included  be- 
tween these  planes,  is  called  a  truncated 
pyramid.  If  the  cross  section  is  parallel  to 
the  base,  the  figure  is  called  a  frustum  of  a  pyramid,  and 
this  section  is  the  upper  base. 

The  altitude  of  a  frustum  is  the  perpendicular  distance 
between  its  bases.  The  slant  height  of  the  frustum  of  a 
regular  right  pyramid  is  the  common  altitude  of  its 
trapezoidal  faces. 


PYRAMIDS  AND  CONES.  77 

145,  THEOREM.  The  volume  of  a  frustum  of  a 
pyramid  is  equal  to  the  combined  volume  of  three 
pyramids  whose  common  altitude  is  the  same  as  that 
of  the  frustum,  and  whose  bases  are  the  upper  and 
lower  bases  of  the  frustum  and  the  mean  proportional 

between  these  bases. 

p 


A  B 

Given  the  frustum  AC  with  lower  base  b,  upper  base  b',  and 
altitude  h.  Let  h'  be  'the  altitude  PM  of  the  completed  pyramid 
P-ABCDE.  Then  V&&'  is  the  mean  proportional  between  b  and 
b'. 

To  prove  that  the  volume  of  ACr  is 


. 

Proof:  The   altitude  of   the   pyramid   P-AfBfCrDfEf    is 
h'  -h. 

Hence'  -'  (§133) 


from  which  A'=  —       --  (1) 

V6-V6' 

Now  V  is  the  difference  between  the  pyramids  whose 
altitudes  are  hf  and  h'  —  h. 

Hence,  V  =  J  bh'  -  \V  (h*  -  A), 

or,  rearranging,  F  =  \lfh  +  \  h'(b  -  b1).    (2) 

Substituting  (1)  in  (2),     F=  J  h[b  +  b'  + 

Show  all  the  details. 


78 


SOLID  GEOMETBY. 


146,  THEOREM.  The  volumes  of  two  tetrahedrons, 
having  a  trihedral  angle  of  the  one  congruent  to  a 
trihedral  angle  of  the  other,  are  in  the  same  ratio  as  the 
products  of  the  edges  which  meet  in  the  vertices  of  these 
angles. 


c' 


Given  the  tetrahedrons  P-ABC  and  P'-A'B'C'  whose  volumes 
are  F  and  V  and  in  which  Tri.  Z  P=  Tri.  Z  P'. 

PA  '  PB  •  PC 


„,  ,  i     ,        v  FA  •  rts  • 

To  prove  that    -—  =  — — :— - 

V'     P'A'  -  P'B'  • 


>of  :  Place  P'-A'B'C'  so  that  Tri.  Z  p'  coincides  with 
Tri.  2" 

Let  CM  and  C'M'  be  the  altitudes  of  P-ABC 
from  the  vertices  C  and  c1  upon  the  plane  PAB. 

Let  ^7V  and  ^i'^'be  the  altitudes  of  the  A  PAB  and  P^'tf'. 

area  PAB  CM  -  PB  •  ^JV 

F'     4  c'jf'  •  area  PA'B'  ~~  C'M'  -  PB'  -  A'N 


Th        V  - 

1       ~ 


Now  prove  — - —  = 
C'jf' 


and 


PA 
PA1 


Hence,  substituting  in  (1), 


,  F  PC  -  PB  •  PA 

we  have  — -  =  — :— : 1—. ; — r  • 

F'     P'C'  •  P'B'  •  P'A' 

Give  all  the  steps  and  reasons  in  detail. 


PYRAMIDS  AND   CONES.  79 

147.  EXERCISES. 

1.  Show  that  the  lateral  faces  of  a  frustum  of  a  regular  pyramid 
are  congruent  isosceles  trapezoids.     Hence  find  the  area  of  its  lateral 
surface  in  terms  of  the  slant  height  and  the  perimeters  of  the  bases. 

2.  Show  that  sections  of  a  pyramid  made  by  two  planes  parallel 
to  the  base  are  similar  polygons  whose  areas  are  in  the  same  ratio  as 
the  squares  of  the  distances  from  the  vertex. 

3.  Show  that  any  two  pyramids  standing  on  the  same  base,  or  on 
equal  bases  in  the  same  plane,  have  the  same  volume  if  their  vertices 
coincide  or  lie  in  a  plane  parallel  to  the  base. 

4.  Show  that  the  volumes  of  two  pyramids  have  the  same  ratio  as 
the  areas  of  their  bases  if  they  have  equal  altitudes,  and  the  same 
ratio  as  their  altitudes  if  they  have  equal  bases. 

5.  A  frustum  of  a  pyramid  is  cut  from  a  pyramid  the  perimeter 
of  whose  base  is  60  inches  and  whose  altitude  is  15  inches.     What  is 
the  altitude  of  the  frustum,  if  the  perimeter  of  its  upper  base  is  40 
inches? 

Does  the  result  depend  upon  the  number  of  sides  of  the  pyramid  ? 

6.  Solve  the  preceding  problem  if  the  perimeter  of  the  upper  base 
of  the  frustum  is  one  nth  that  of  the  lower  base.     Does  this  result 
depend  upon  the  number  of  sides  of  the  pyramid  ? 

7.  The  area  of  the  base  of  a  pyramid  is  180  square  inches  and  its 
altitude   is  20  inches.     Cut  from  it  a  frustum,  the  area  of  whose 
upper  base  is  45  square  inches ;  also  one  the  area  of  whose  upper  base 
is  one  nth  of  180  square  inches.     Do  these  results  depend  upon  the 
number  of  sides  of  the  pyramid? 

8.  Two  triangular  pyramids  have  equal  trihedral  angles  at  the 
vertex.     The  lateral  edges  of  one  pyramid  are  14,  16,  and  18,  and 
those  of  the  other  7,  8,  and  9.     Find  the  ratio  between  their  volumes. 
Are  the  data  given  sufficient  to  find  the  volume  of  each  pyramid? 

9.  If  two  triangular  pyramids  have  equal  trihedral  angles  at  the 
vertex  and  if  the  lateral  edges  of  one  are  a,  &,  and  c,  and  two  lateral 
edges  of  the  others  are  a'  and  b',  find  the  third  lateral  edge  of  the 
second  pyramid  so  that  their  volumes  shall  be  equal. 

10.  The  slant  height  of  a  frustum  of  a  regular  pyramid  is  10 
inches  and  the  apothems  of  its  bases  8  and  6  inches  respectively. 
Find  its  altitude. 


80 


SOLID  GEOMETRY. 


CONES. 

148,  Definition.  Given  a  closed  convex 
curve  and  a  fixed  point  not  in  its  plane.  If 
a  line  through  the  fixed  point  moves  so  as 
always  to  touch  the  curve  and  is  made  to 
traverse  it  completely,  it  is  said  to  generate 
a  convex  conical  surface. 

The  moving  line  is  called  the  generator" 
of  the  surface,  and  in  any  particular  posi- 
tion it  is  an  element  of  the  surface. 

The  fixed  curve  is  called  the  directrix, 
and  the  fixed  point  the  vertex. 

A  conical  surface  consists  of  two  parts, 
called  nappes,  on  opposite  sides  of  the  fixed 
point. 


149,  That  part  of  a  convex  conical  sur- 
face included  between  its  vertex  and  a 
plane  cutting  all  its  elements,  together 
with  the  intercepted  portion  of  the  plane, 
is  called  a  cone. 

The  intercepted  part  of  the  plane  is  the  base  of  the 
cone,  and  the  curved  surface  is  its  lateral  surface. 

The  altitude  of  a  cone  is  the  perpen- 
dicular distance  from  the  vertex  to  the  plane 
of  the  base. 

A  circular  cone  is  one  which  has  a  cir- 
cular cross  section   such  that   the  perpen- 
dicular upon  it  from  the  vertex  meets  it  at 
the  center.     If  the  base  is  such  a  circle,  the     c 
cone  is  then  called  a  right  circular  cone.     Otherwise,  it 
is  an  oblique  circular  cone. 


PYRAMIDS  AND   CONES. 


81 


A  right  circular  cone  may  be  generated 
by  rotating  a  right  triangle 
PMB  about  one  of  its  legs, 
P.M,  as  an  axis.  The  hypot- 
enuse PB  generates  the  con- 
ical surface,  and  the  other 
leg,  MB,  generates  the  base. 

The  generator  of  a  right  circular  cone  in 
any  position  is  called  the  slant  height. 

150,  THEOREM.  If  a  plane  contains  an  element  of 
a  cone  and  meets  it  in  one  other  pointy  then  it  contains 
another  element  also,  and  the  section  is  a  triangle. 


Let  a  plane  contain  the  element  PD  of  the  cone  P-ABC,  and 
also  one  other  point  B. 

To  prove  that  this  plane  contains  another  element  PB, 
and  that  the  section  is  a  triangle  PBD. 

Suggestion.  Connect  P  and  B.  This  segment  lies  in 
the  conical  surface.  (Why  ?)  Complete  the  proof  by 
showing  that  BD  lies  in  the  base  of  the  cone.  Compare 
this  proof  with  that  of  §  100. 

151.  Definition.  If  a  plane  contains  an  element  of  a 
cone  and  no  other  point  of  the  cone,  the  plane  is  tangent  to 
the  cone,  and  the  element  is  called  the  element  of  contact. 


82  SOLID   GEOMETRY. 

152,  THEOREM.    If  the  base  of  a  cone  is  circular,  every 
plane  section  parallel  to  the  base  is  also  circular. 


c 

Given  a  cone  with  a  circular  base  AD. 

To  prove  that  the  H  section  EH  is  also  circular. 

Proof :  Draw  the  straight  line  from  p  to  the  center  M 
of  the  base,  and  let  it  meet  the  section  EH  in  the  point  o. 
Let  F  and  G  be  any  two  points  on  the  perimeter  of  the 
section  EH. 

Pass  planes  containing  PM  through  the  points  F  and  G, 
and  let  them  cut  the  base  in  MB  and  MC  respectively. 

Now  in  the  A  PMB  and  PMC  prove  that  OF=  OG. 

Hence,  as  F  and  G,  any  two  points  on  the  perimeter  of 
this  section,  are  equally  distant  from  O,  this  shows  that 
EH  is  a  circle  whose  center  is  o. 

153,  COROLLARY.  If  a  cone  has  a  circular  base,  the 
areas  of  two  parallel  cross  sections  are  in  the  same 
ratio  as  the  squares  of  their  perpendicular  distances 
from  the  vertex  and  also  as  the  squares  of  the  distances 
of  their  centers  from  the  vertex. 

Suggestion.  Use  the  figure  of  §  152,  and  let  PQ  be  the 
altitude  of  the  cone.  Then  show  that 

Area  AD  _  3fZ)2  _  PM2  _  PQ2 
Area  EH 


P  YE  AMID  S  AND   CONES.  83 

154.  EXERCISES. 

1.  Into  how  many  parts  do  the  two  nappes  of  a  conical  surface  or 
of  a  pyramidal  surface  divide  the  remaining  points  of  space  ? 

2.  If  in  constructing  a  conical  surface  a  polygon  is  used  as  a 
directrix  instead  of  a  closed  convex  curve,  what  kind  of  surface  is 
obtained  ? 

3.  Why  is  it  specified  in  the  definition  of  the  convex  conical  sur- 
face that  the  vertex  must  not  lie  in  the  same  plane  as  the  directrix? 

4.  How  many  cones  may  be  cut  from  a  conical  surface  if  they  are 
to  have  no  point  in  common  except  the  vertex? 

5.  If  a  triangle  which  is  not  a  right  triangle  is  made  to  revolve 
about  one  of  its  sides,  does  it  generate  a  cone? 

6.  Can  every  circular  cone  be  developed  by  revolving  a  right  angled 
triangle  about  one  of  its  sides  ? 

7.  If  a  cone  has  a  circular  base,  the  line  from  the  vertex  to  the 
center  passes  through  the  center  of  every  plane  section  parallel  to  the 
base. 

8.  If  a  cone  has  a  circular  base,  the  plane  determined  by  a  tangent 
to  the  base  and  the  element  at  the  point  of  tangency  is  a  tangent  plane 
to  the  cone. 

9.  Through   a  point   outside   a  cone  with  a  circular  base,  how 
many  planes  are  there  which  are  tangent  to  the  cone? 

10.  The  diameter  of  the  circular  base  of  a  cone  is  8  in.  and  the 
altitude  of  the  cone  9  in.     A  plane  parallel  to  the  base  cuts  the  cone 
in  a  section  whose  diameter  is  3  in.     Find  the  distance  from  the 
vertex  of  the  cone  to  this  plane. 

11.  If  the  area  of  the  circular  base  of  a  cone  is  16  TT  sq.  in.  and  its 
altitude  6  in.,  find  the  distance  from  the  vertex  to  a  plane,  parallel  to 
the  base,  which  cuts  the  cone  in  a  section  with  area  9  TT  sq.  in. 

12.  The  area  of  the  circular  base  of  a  cone  is  b  sq.  in.  and  its 
altitude  h  in.     Find  the  distance  from  the  base  to  a  plane  parallel  to 
it,  which  cuts  off  a  cone  the  area  of  whose  base  is  one  nth  that  of  the 
base  of  the  original  cone. 

13.  Compare  the  exercises  on  the  cone  thus  far  studied  with  those 
on  the  pyramid  given  on  pages  76,  79.     Note  that  a  pyramid  can  be 
made  to  approximate  very  closely  to  a  cone  by  making  its  number  of 
faces  very  large. 


84  SOLID   GEOMETRY. 


MEASUREMENT   OF   THE   SURFACE  AND  VOLUME  OF  A   CONE. 

155,  Definitions.     A  pyramid  is  said  to  be  inscribed  in  a 
cone  if  its  lateral  edges  are 

elements  of  the  cone,  and 
the  bases  of  the  cone  and 
the  pyramid  lie  in  the  same 
plane,  as  in  Fig.  2. 

A  pyramid  is  said  to  be 
circumscribed  about  a  cone 
if  its  lateral  faces  are  all 

-tangent  to  the   cone,  and  the  bases  of  the  cone  and  the 
pyramid  lie  in  the  same  plane,  as  in  Fig.  1. 

156,  THEOREM.     In  a  right  circular  cone  a  pyramid 
may  be  inscribed  whose  slant  height  differs  from  the 
slant  height  of  the  cone  by  less  than  any  given  fixed 
number. 

Proof  :  Let  d  be  the  given  fixed  number.  From  plane 
geometry  we  know  that  a  regular  polygon  may  be  inscribed 
in  the  base  of  the  cone  whose  apothem  differs  from  the 
radius  of  the  base  by  less  than  d.  Let  this  regular  poly- 
gon be  the  base  of  an  inscribed  pyramid.  Then  the  slant 
height  of  this  pyramid  differs  from  that  of  the  cone  by  less 
than  d,  since  the  difference  of  two  sides  of  a  triangle  is  less 
than  the  third  side. 

157,  Axiom  VIII.      The  lateral  surface  of  a  convex 
cone  has  a  definite  area,  and  the  cone  incloses  a  definite 
volume,  which  are  less  respectively  than  those  of  any 
circumscribed  pyramid  and  greater  than  those  of  any 
inscribed  pyramid. 


•  *  \ 

PYRAMIDS  AND   CONES.    .  85 

158,  THEOREM.     The  area  of  the  lateral  surface  of  a 
right  circular  cone  is  equal  to  one  half  the  product  of 
its  slant  height  and  the  circumference  of  its  base. 

Given  a  right  circular  cone  of  which  s  is  the  slant  height,  c  is 
the  circumference  of  the  base,  and  L  the  lateral  area. 

To  prove  that  L  =  J  so. 

Proof:  Suppose  that  L>^sc.  Then  L  =  %  sK  (1) 
where  K  >  c. 

Circumscribe  about  the  cone  a  pyramid  the  perimeter  of 
whose  base  is  p,  such  that  p<K.  (Why  is  this  possible  ?) 

Hence,  -|  sp  <  J  SK.  (2) 

That  is,  (2)  contradicts  (1)  because  of  §  157. 
Next  suppose  L  <  |  so.     Then  L  =  \  sKr      (3) 

where  K'  <  c. 

Let  c  —  K^  —  d.  By  §  352,  Plane  Geometry,  a  polygon 
may  be  inscribed  in  the  base  of  the  cone  whose  perimeter 
p  differs  from  c  by  as  little  as  we  please,  and  by  §  156  a 
pyramid  may  be  inscribed  in  the  cone  whose  slant  height  sf 
differs  from  the  slant  height  s  of  the  cone  by  as  little  as  we 
please.  Hence  a  pyramid  may  be  inscribed  such  that 
J  s'p  differs  from  J  so  by  less  than  d. 

That  is  is'p>±sK.  (4) 

But  (4)  contradicts  (3)  because  of  §  157. 
Since  therefore  L  is  neither  less  than  nor  greater  than 
^  s<?,  it  must  be  equal  to  J  sc. 

159,  COROLLARY.     If  r  is  the  radius  of  the  base  of 
a  right  circular  cone  and  s  the  slant  height,  then 

L  =       2  irrS  =  irrS. 


86 


SOLID   GEOMETRY. 


160,  Definition.     Two  right  circular  cones  are  similar  if 
they  are  generated  by  two  similar  right  triangles  revolving 
about  corresponding  sides. 

161,  THEOREM.     The  lateral  areas  or  the  entire  areas 
of  two  similar  right  circular  cones  are  in  the  same  ratio 
as  the  squares  of  their  altitudes,  their  slant  heights,  or 
the  radii  of  their  bases. 


Given  the  cones  P  and  P',  with  altitudes  h,  h',  slant  heights  s,  s', 
radii  of  bases  r,  r'}  lateral  areas  L,  L',  and  entire  areas  A,  A'- 

To  prove  that 

4L=LL=  W=s2L=r*_ 
A1      L'      A'2      s'2     r'2' 

Proof:  See  the  suggestions  under  §  111.     Give  all  the 
steps. 

162.  EXERCISES. 

1.  The  lateral  area  of  a  cone  is  36  square  inches.     What  is  the 
lateral  area  of  a  similar  cone  whose  altitude  is  f  that  of  the  given 
cone? 

2.  The  total  area  of  one  of  two  similar  cones  is  three  times  that  of 
the  other.     Compare  their  altitudes  and  also  their  radii. 

3.  The  sum  of  the  total  areas  of  two  similar  cones  is  144  square 
inches.     Find  the  area  of  each  cone  if  one  is  1|  times  as  high  as  the 
other. 


PYRAMIDS  AND   CONES.  87 

4.  Prove  that  if  in  two  tetrahedrons  three  faces  of  one  are  congru- 
ent respectively  to  three  faces  of  the  other  and  similarly  placed  about 
a  vertex,  the  tetrahedrons  are  congruent. 

5.  Prove  that  if  in  two  tetrahedrons  two  faces  and  the  included 
dihedral  angle  are  congruent  and  similarly  placed,  the  tetrahedrons 
are  congruent. 

6.  A  pedestal  for  a  monument  is  in  the  shape  of  a  frustum  of  a 
regular  hexagonal  pyramid,  the  radius  of  the  upper  base  being  4  ft., 
that  of  the  lower  base  6  ft.,   and  the  altitude  of  the  frustum  8  ft. 
Find  its  volume,  slant  height,  and  lateral  surface. 

7.  Find  the  volume  of  a  frustum  of  a  pyramid  the  areas  of  whose 
bases  are  25  sq.  in.  and  18  sq.  in.  and  and  whose  altitude  is  6  in. 

8.  The  area  of  the  lower  base  of  a  frustum  is  42  sq.  ft.,  its  alti- 
tude 8  ft.,  and  volume  200  cu.  ft.     Find  the  area  of  the  upper  base. 

9.  The  area  of  the  base  of  a  pyramid  is  480  sq.  ft.  and  its  altitude 
30  ft.     Find  the  volume  of  the  frustum  remaining  after  a  pyramid 
with  altitude  10  ft.  has  been  cut  off  by  a  plane  parallel  to  the  base. 

10.  The  area  of  the  base  of  a  pyramid  is  250  sq.  in.     If  a  plane 
section  of  the  pyramid  parallel  to  the  base  and  at  a  distance  of  5  in. 
from  it  has  an  area  of  175  sq.  in.,  find  the  altitude  of  the  pyramid. 

11.  The  sides  of  the  base  of  a  triangular  pyramid  are  6  ft.,  8  ft., 
10  ft.,  and  its  volume  96  cu.  ft.     Find  its  altitude. 

12.  What  part  of  the  volume  of  a  cube  is  a  frustum  of  a  pyramid 
cut  from  a  pyramid  whose  base  is  one  face  of  the  cube  and  whose 
vertex  lies  in  the  opposite  face,  if  the  altitude  of  the  frustum  is  one 
half  the  edge  of  the  cube  ? 

13.  Find  the  dihedral  angle  at  the  base  of  a  regular  pyramid  if 
the  altitude  is  one  half  the  slant  height. 

14.  A  right  triangular  prism  is  cut  by  a  plane 
not  parallel  to  the  base,  but  such  that  its  inter- 
section DE  is  parallel  to  the  base  segment  AB. 
Show  that  the  volume  of  the  part  thus  cut  off 
is  one  third  the  product  of  the  sum  of  the  three 
vertical  edges  and  the  area  of  the  base. 

SUGGESTION.     Draw  plane  DEK  II  ABC. 

15.  Find  the  volume  of  a  truncated  triangular  prism,  the-  area  of 
whose  base  is  25  square  inches  and  whose  lateral  edges  are  8,  7,  8. 


88 


SOLID   GEOMETRY. 


163.  THEOREM.  The  lateral  area  of  a  frustum  of 
a  right  circular  cone  is  equal  to  one  half  of  the  sum 
of  the  circumferences  of  the  bases  multiplied  by  the 
slant  height.  P 


Given  the  frustum  ABCD,  with  slant  height  s  and  radii  r  and 
r'.     Let  L  represent  its  lateral  area. 

To  prove  that  L  =  |-  (2  irr  +  2  TH*')  s  =  TTS  (r  +  r'). 
Proof  :  Complete  the  cone,  and  let  PC  =  sf . 
Then  L  =  $  [2  TTT  (s  +  «')  -  2 

=  Trrs  +  TTS'  (r  —  /). 


f  ,  .  ,     , 

-,  irom  wmcn  s   = 


But  S.-'S^r   , 

A.'  O'  A.  /]»' 

Substituting  s'  from  (2)  in  (1), 

L  =  TTTS  -f-  TTT'S  =  TTS  (r  +  f' )  • 


a) 

(2) 


164,  COROLLARY.  77ie  lateral  area  of  a  frustum 
of  a  right  circular  cone  is  equal  to  the  circumference  of 
a  section  midway  between  the  bases  multiplied  by  the 
slant  height. 

Suggestion.     From  the  theorem 

L  =  7r3(T  .  rf)  =  <>7r(r  +  r'')3 

2 

Now  show  that  r          is  the  radius  of  the  section  mid- 


way  between  the  two  bases. 


- 

PYRAMIDS  AND  CONES.  89 


165,  EXERCISES. 

1.  The  lateral  surface  of  a  right  circular  cone  is  75  sq.  ft.     Find 
the  altitude  if  the  radius  of  the  base  is  4  ft. 

2.  A  circular  chimney  100  ft.  high  is  in  the  form  of  a  frustum  of 
a  right  cone  whose  lower  base  is  10  ft.  in  diameter  and  upper  base 
8  ft.     Find  the  lateral  surface. 

3.  The  lateral  area  of  a  frustum  of  a  right  circular  cone  is  60  TT 
sq.  in.,  the  radii  of  the  two  bases  are  6  in.  and  4  in.     Find  the  slant 
height  of  the  frustum. 

4.  Find  the  altitude  of  the  frustum  in  the  preceding  example,  and 
also  the  altitude  of  the  cone  from  which  it  is  cut. 

5.  A  frustum  of   a  right  circular  cone  has  an  altitude  one  half 
that  of  the  cone  from  which  it  is  cut.     If  its  slant  height  is  8  ft.  and 

lateral  area  64  TT  sq.  ft.,  find  the  diameters  of  its  bases. 

/ 

6.  Find  the  altitude  of  the  frustum  of  the  cone  in  the  preceding 

example ;  also  the  lateral  area  of  the  cone  from  which  it  was  cut. 

166,    THEOREM.      The  volume  of  any  convex  cone  is 
equal  to  one  third  the  product  of  its  base  and  altitude. 


Suggestion.  Let  h  be  the  altitude,  b  the  area  of  the 
base,  and  v  the.  volume. 

Show  that  V  cannot  be  different  from  J  bh  by  an  argu- 
ment similar  to  that  of  §  108,  making  use  of  §  141. 

167,  COROLLARY.  If  a  cohe  has  a  circular  base  of 
radius  r  and  altitude  h,  then  ^=  ^TrrA. 


90 


SOLID   GEOMETRY. 


168,  THEOREM.  The  volume  of 
the  frustum  of  a  convex  cone  is 
equal  to  the  combined  volumes  of 
three  cones  whose  comm  on  altitude 
is  the  altitude  of  the  frustum  and 
whose  bases  are  the  upper  and 
lower  bases  of  the  frustum  and  a 
mean  proportional  between  these 
bases. 

Suggestion.  The  proof  is  ex- 
actly like  that  of  §  145. 

NOTE.  Observe  that  the  two  preceding  theorems  apply  to  any  con- 
vex cone  because  the  altitude  h  is  constant.  The  actual  computation 
is  practicable  only  when  the  areas  of  the  bases  can  be  found,  as  in  the 
case  of  the  circle,  or  ellipse. 

SUMMARY  OF   CHAPTER  III. 

1.  Make  a  list  of  definitions  on  pyramids  and  also  one  on  cones  and 
compare  them. 

2.  Make  a  list  of  theorems  on  pyramids  and  also  one  on  cones  and 
compare  them. 

3.  What  axioms  have  been  used  in  this  chapter  ?     Compare  these 
with  the  axioms  in  Chapter  II. 

4.  Make  a  list  of  all  the  formulas  given  by  the  theorems  of  this 
chapter  and  compare  them  with  the  corresponding  formulas  in  Chap- 
ter II. 

5.  What  theorems  on  cylinders  have  no  corresponding  theorems 
for  cones  ? 

6.  Show  that  a  frustum  of  a  cone  becomes  more  and  more  nearly 
identical  with  a  cylinder  if  the  vertex  of  the  cone  is  removed  farther 
and  farther  from  the  base. 

7.  Make  a  list  of  the  applicajbions  in  this  chapter  which  have  im- 
pressed you  as  interesting  or  practical  or  both.     Return  to  this  ques- 
tion again  after  studying  the  problems  and  applications  which  follow. 


PYRAMIDS  AND  CONES. 


91 


PROBLEMS  AND  APPLICATIONS. 

1.  Show  that  the  volume  of  a  right  triangular  prism  is  equal  to 
one  half  the  product  of  the  area  of  one  face  and  the  distance  from  the 
opposite  edge  to  that  face. 

2.  A  mound  of  earth  in  the  shape  shown  in  the  figure  has  a  rec- 
tangular base  16  yards  long  and  8  yards  wide. 

Its  perpendicular  height  is  5  yards,  and  the 
length  on  top  is  8  yards.  Find  the  number 
of  cubic  yards  of  earth  in  the  mound. 

SUGGESTION.     If  from  each  end  a  pyramid  484 

with  a  base  8  yd.  by  4  yd.  is  removed,  the  remaining  part  is  a  tri- 
angular prism. 

3.  Given  a  figure  in  general  shape  the  same  as  the  preceding,  with  a 
rectangular  base  of  length  24  ft.  and  width  6  ft.     Find  its  volume  and 
lateral  area  if  the  dihedral  angles  around  the  base  are  each  45°. 

4.  The   accompanying  figure  represents  a  solid  whose  base  is   a 


15 


15 


rectangle  50  units  long  and  40  units  wide.     Its  height  is  12  units  and 
its  top  a  rectangle  20  units  by  10  units.     Find  its  volume. 

SUGGESTION.  Divide  the  solid  as  indicated 
in  the  figure.  Notice  that  this  is  not  a  frus- 
tum of  a  pyramid. 

5.  In  a  figure  like  the  foregoing,  how  can 
we  determine  whether  or  not  it  represents  the 
frustum  of  a  pyramid  ? 

6.  Show   how  to  pass  a  plane  through  a 
tetrahedron   so   that  the  section   shall   be  a 
parallelogram. 

SUGGESTION.  Pass  a  plane  parallel  to  each 
of  two  opposite  edges.  See  Ex.  6,  page  15. 


92 


SOLID  GEOMETRY. 


7.  If  the  middle  points  of  four  edges  of  a  tetrahedron,  no  three 
of  which  meet  at  the  same  vertex,  are  joined,  a  parallelogram  is 
formed.  Prove. 


K 

8.  If  through  any  point  P  in  a  diagonal  of  a  parallelepiped  planes 
KN  and  RM  are  drawn  parallel  to  two  faces,  show  that  the  parallele- 
pipeds DQ,  and  LN  thus  formed  have  equal  volumes. 

9.  Find  the  volume  and  area  of  a  figure  formed  by  revolving  an 
equilateral  triangle  about  an  altitude,  the  sides  of  the  triangle  being  s. 

10.  Find  the   area   and  volume  of  the  figure   developed   by  an 
equilateral  triangle  with  sides  s  if  it  is  revolved 

about  one  of  its  sides. 

11.  Find   the   area   and  volume  of  the  figure 
developed  by  revolving  a  square  whose   side  is  s 
about  one  of  its  diagonals. 

12.  Through  one  vertex  of  an  equilateral  tri- 
angle with  sides  s  draw  a  line  I  perpendicular  to 
the  altitude  upon  the  opposite  side.     Find  the  vol- 
ume and  area  of  the  figure  developed  by  revolving 
the  triangle  about  the  line  I. 

13.  Through  a  vertex  of  a  square  with  sides  s 

draw  a  line  I  perpendicular  to  the 
diagonal  through  that  vertex.  Find 
the  area  and  volume  of  the  figure 
developed  by  turning  the  square 
around  the  line  /. 


14.  In  a  regular  hexagon  with  sides  s  draw  a 
line  I  bisecting  two  opposite  sides.  Find  the  area 
and  volume  of  the  figure  developed  by  turning  the 
hexagon  about  I  as  an  axis. 


JV 

PYRAMIDS  AND  CONES.  93 

15.  Solve  a  problem  like  the  preceding,  using  a  regular  octagon 
instead  of  a  hexagon. 

16.  If  several  planes  are  tangent  to  the  same 
cone,  find  one  point  common  to  them  all. 

17.  Find  the  locus  of  all  lines  which  make  a 
given  angle  with  a  given  line  at  a  given   point 
in  it. 

18.  Find  the  locus  of  all  lines  which  make  a 
given  angle  with  a  given  plane  at  a  given  point. 

19.  One  angle  of  a  right  triangle  is  30°.     Find 

the  ratios  between  the  surfaces  of  the  solids  developed  by  revolving 
this  triangle  around  each  of  its  three  sides. 

20.  Find  the  ratios  between  the  volumes  of  the  solids  developed  in 
the  preceding  example. 

21.  Find  the  total  area  and  the  volume  of  a  regular  tetrahedron 
each  of  whose  edges  is  e.     (See  §  170.) 

22.  If  the  numerical  values  of  the  volume  and  of  the  total  area  of 
a  regular  tetrahedron  are  equal,  what  is  the  length  of  its  edge  ? 

23.  Find  the  length  of  an  edge  of  a  regular  tetrahedron  if  its 
volume  is  numerically  equal  to  the  square  of  an  edge. 

24.  Cut  a  pyramid  of  altitude  h  by  means  of  a  plane  parallel  to 
the  base  so  that  the  perimeter  of  the  section  shall  be  half  that  of  the 
base.      Also  cut  it  so  that  the  perimeter  of  the  section  shall  be  £ 
that  of  the  base. 

25.  Cut  a  right  circular  cone  by  3  planes,  each  parallel  to  the  base, 
so  that  the  perimeters  of  the  sections  shall  be  ^,  ^,  — j£,  p  being 

the  perimeter  of  the  base.     Find  the  distances  from  the  vertex  to  the 
planes. 

26.  Cut  a  right  circular  cone  of  altitude  h  by  a  plane  parallel  to 
the  base  so  that  the  area  of  the  section  shall  be  half  that  of  the  base. 
Find  the  distance  from  the  vertex  to  the  plane. 

27.  Show  that  the  lateral  area  of  the  small  cone  cut  off  in  the 
preceding  example  is  one  half  the  lateral  area  of  the  original  cone. 

28.  Cut  a  pyramid  of  altitude  h  by  n  planes,  each  parallel  to  the 

base,  so  that  the  areas  of  the  sections  shall  be , -,  ..., 

n  +  l'n  +  1    n+l 


94 


SOLID   GEOMETRY. 


A  being  the  area  of  the  base. 

1 


that  the  distances 
3 


-allel  to  the  base  so 
ial  half  that  of  the 
me  to  the  plane, 
•arallel  to  the  base, 
cut  off  at  the  top 


(n-l)A 

n+l 

from  the  vertex  to  the  planes  are  h  v , 

*n  -f  1 

29.  Cut  a  cone  with  altitude  h  by  a  plane 
that  the  volume  of  the  frustum  formed  shall 
cone.     Find  the  distance  from  the  vertex  of  the 

30.  Cut  a  cone  of  altitude  h  by  n  planes,  each 
so  that  the  frustums  formed  and  the  one 

shall  all  have  equal  volumes. 

31.  An  equilateral   triangle  ABC  is  swung 
around  the  line  DE  as  an  axis,  D  and  E  being 
middle  points  of  the  sides  of  the  triangle.     Find 
the  volume  of  the  figure  thus  developed  by  the 
trapezoid  ABED  if  AB  =  a. 

32.  Find  the  total  surface  of  the  figure  in 
the  preceding  example. 

33.  In  a  right  circular  cone,  with  altitude  h, 
and  r  the  radius  of  its  base,  a  cylinder  is  inscribed 
as  shown  in  the  figure.     Find  the  radius  OF  of 
the  cylinder  if  the  area  of  the  ring  bounded  by 
the  circles  OF  and  OA  is  equal  to  the  lateral 
area  of  the  small  cone  cut  off  by  the  upper  base 
of  the  cylinder. 

34.  The  same  as  the  preceding,  except  that  the  lateral  area  of  the 
small  cone  is  to  equal  the  lateral  area  of  the  cylinder. 

35.  Find  the  dihedral  angles  of  a  regular  tetrahedron. 


CHAPTER  IV. 

REGULAR  AND  SIMILAR  POLYHEDRONS. 
REGULAR  POLYHEDRONS. 

169,  Definitions.     A  polyhedron  is  said  to  be  regular  if 
its  polyhedral  angles  are  all  congruent  and  its  faces  are 
congruent  regular  polygons. 

170,  Construction  of  regular  polyhedrons.     Certain  regu- 
lar  polyhedrons   are    very    simple    of 

construction,  as  indicated  below. 

(1)  The  regular   tetrahedron.      At 
the   center   E  of    an    equilateral   tri- 
angle  ABC  erect  a  perpendicular   to 
the  plane  of   the  triangle.     On   this 
take  a  point  D  so  that  AD  =  AC. 

Now  prove  that  the  four  triangles, 
ABC,  ACD,  ABD,  BCD,  are  regular  and 
congruent,  and  that  the  four  trihedral 
angles  are  congruent. 

Suggestion.     AE  =  BE  =  CE. 

(2)  The  regular  hexahedron  or  cube. 
At  the  vertices  of  a  given  square  erect 
perpendiculars   to   its  plane  equal  in 
length  to  the  sides,  and  join  their  up- 
per extremities  as  shown  in  the  figure. 

Show  that  six  equal  and  congruent  squares  are  formed 
and  also  eight  congruent  trihedral  angles. 

95 


96 


SOLID   GEOMETRY. 


(3)  The  regular  octahedron.     Through  the  center  o  of 
a  square  ABCD  draw  a  perpendicular 

to  the  plane  of  the  square. 

On  this  take  points  E  and  F  such 
that  AF=AE  =  AB.  Join  E  and  F  to 
each  of  the  four  vertices,  A,  J3,  C,  D. 

Now  prove  that  the  eight  faces  are 
congruent  regular  triangles,  and  that 
the  six  polyhedral  angles  are  con- 
gruent. 

There  are  two  other  regular  polyhe- 
drons, namely : 

(4)  The  regular  dodecahedron,  having 
for  faces  twelve  regular  pentagons,  and 


TETRAHEDRON 


HEXAHEDRON 


OCTAHEDRON 


DODECAHEDRON 


ICOSAHEDRON 


REGULAR  AND   SIMILAR  POLYHEDRONS.  97 


(5)  The  regular  icosahedron,  having 
for  faces  twenty  equilateral  triangles. 

The  geometric  constructions  for  (4) 
and  (5)  are  not  so  simple  as  for  the  others. 

However,  cardboard  models  of  all  five 
may  be  made  by  cutting  out  the  figures, 
as  shown  herewith,  and  folding  them  along  the  dotted 
lines.  They  may  be  held  in  shape  by  means  of  gum  paper 
stuck  over  the  joints. 

171,  The  number  of  regular  polyhedrons.  It  will  now 
be  shown  that  there  are  not  more  than  these  five  regular 
polyhedrons. 

There  must  be  at  least  three  faces  meeting  at  each 
vertex.  If  these  are  regular  triangles,  there  may  be  three, 
as  in  the  tetrahedron,  or  four,  as  in  the  octahedron,  or 
five,  as  in  the  icosahedron ;  but  there  cannot  be  six,  for  in 
that  case  the  sum  of  the  angles  about  a  vertex  would  be 
360°,  and  it  is  readily  seen  that  this  sum  cannot  be  as 
great  as  360°.  For  a  proof  of  this  fact,  see  §  271. 

If  the  faces  are  squares,  there  may  be  three  about  a 
vertex,  as  in  the  cube,  but  there  cannot  be  four,  for  in  that 
case  the  sum  of  the  face  angles  at  a  vertex  would  be  360°. 

If  the  faces  are  regular  pentagons,  there  may  be  three 
about  a  vertex,  making  the  sum  of  the  face  angles 
3  •  108°  =  324°,  but  there  cannot  be  four,  for  then  the  sum 
would  be  greater  than  360°.  See  §  271. 

Regular  polygons  of  more  than  five  sides  cannot  form 
the  faces  of  a  regular  polyhedron,  for  the  sum  of  the  face 
angles  at  a  vertex  would  in  any  such  case  be  more  than 
360°.  Show  why  this  is  so. 

Hence,  there  cannot  be  more  regular  convex  polyhedrons 
than  those  exhibited  above. 


98  SOLID  GEOMETRY. 

172,  THEOREM.     If  v  is  the  number  of  vertices  of  a 
convex  polyhedron,  e  its  number  of  edges,  and  f  its 
number  of  faces,  then  e  +  2  =  v+f. 

This  is  called  Euler's  Theorem.  The  proof  is  too  dif- 
ficult for  an  elementary  text-book  such  as  this.  The 
proofs  given  in  the  current  texts  are  not  conclusive. 

173,  EXERCISES. 

1.  Verify  the  above  theorem  by  counting  the  number  of  edges, 
faces,  and  vertices  in  each  of  the  regular  figures  given  in  §  170. 

2.  The  following  is  a  form  of  proof  of  this  theorem  which  is  often 
given : 

Denote  the  number  of  vertices,  edges,  and  faces  of  a  polyhedron 
by  F,  E,  and  F,  respectively.  To  prove  E  +  2  =  V+F. 

Proof:  Taking  the  single  face  ABCD,  the  number  of  edges  equals 
the  number  of  vertices,  or  E.  =  V.  If  another  face  be  annexed,  three 
new  edges  and  two  new  vertices  are  added.  Hence  the  number  of 
edges  gains  one  on  the  number  of  vertices,  as  E  —  F+  1.  If  still 
another  face  be  added,  two  new  edges  and  one  new  vertex  are  added. 
Hence  E=  F+ 2. 

With  each  new  face  that  is  annexed  the  number  of  edges  gains  one  on 
the  number  of  vertices,  till  but  one  face  is  lacking. 

The  last  face  increases  neither  the  number  of  edges  nor  vertices. 
Hence,  etc. 

Show  by  putting  together  the  faces  of  a  cube  in  a  certain  order 
that  the  statement  in  italics  need  not  be  true,  and  hence  that  the 
proof  is  not  conclusive. 

174,  THEOREM.     The  sum  of  the  face  angles  of  any 
convex  polyhedron  is  equal  to  fouv  times  as  many  right 
angles,  less  eight,  as  the  polyhedron  has  vertices. 

The  proof  depends  on  the  preceding  theorem  and  is 
not  given  here. 

175,  EXERCISE. 

Verify  this  theorem  by  an  examination  of  the  regular  polyhedrons. 


REGULAR  AND  SIMILAR  POLYHEDRONS. 


99 


INSCRIPTION  OF  REGULAR  POLYHEDRONS. 

176,  PROBLEM.     To  find    a  point    equally   distant 
from  the  four  vertices  of  any  tetrahedron. 


Given  the  tetrahedron  P-ABC. 

To  find  a  point  O  equidistant  from  P,  A,  B,  C. 

Construction.  At  I),  the  middle  point  of  BC,  construct  a 
plane  perpendicular  to  BC. 

The  plane  will  contain  the  point  E,  the  center  of  the 
i  circle  circumscribed  about  A  PBC  and  also  the  similar 
point  F  in  A  ABC.  (Why-?) 

In  the  plane  DEF  draw  EG  _L  ED  and  FH  _L  FD. 

Then  EG  and  FH  cannot  be  parallel  (why  ?),  and  hence 
meet  in  some  point  O. 

Also  EG  J_  to  plane  PBC  and  FH  _L  plane  ABC.    (Why  ?) 

Proof :    Now  show  that  O  is  equidistant  from  P,  A,  U,  C. 

177,  Definitions.  The  locus  of  all  points  in  space  equi- 
distant from  a  given  fixed  point  is  a  surface  called  a 
sphere.  The  fixed  point  is  the  center  of  the  sphere,  and 
a  line-segment  from  the  center  to  the  surface  is  called  a 
radius. 

A  polyhedron  is  inscribed  in  a  sphere  if  all  its  vertices 
lie  in  the  sphere. 

The  sphere  is  also  said  to  be  circumscribed  about  the 
polyhedron. 


100 


SOLID   GEOMETRY. 


178, 


EXERCISES. 


1.  In  the  construction  of  §  176  show  that  0  is  the  only  point  equi- 
distant from  P,  A,  B,  and  C. 

2.  Show  that  the  planes  perpendicular  to  each  of  the  six  edges  of 
the  tetrahedron  at  their  middle  points  meet  in  the  point  O. 

3.  Does  the   construction  of  §  176  depend  upon  the  tetrahedron 
being  regular  f     Can  a  sphere  be  circumscribed  about  any  tetrahedron  ? 

4.  Is  there  any  limitation  on  the  relative  position  of  four  points  in 
order  that  a  sphere  may  be  passed  through  them  ? 

179.  Four  points  not  all  lying  in  the  same  plane  are  said 
to  determine  a  sphere. 

Any  other  point,  taken  at  random,  will  not,  in  general, 
lie  on  a  sphere  determined  by  four  given  points. 

Hence,  while  any  tetrahedron  may  be  inscribed  in  a 
sphere,  a  polyhedron,  in  general,  cannot. 

However,  any  regular  polyhedron  may  be  inscribed  in  a 
sphere. 

180,  PROBLEM.     To  inscribe  a  cube  in  a  sphere. 


A  B 

Suggestion.  Show  that  all  the  diagonals  meet  in  a  com- 
mon point  O  which  is  equally  distant  from  all  the  vertices. 

Or  show  that  a  perpendicular  to  one  face  at  its  center  L 
meets  the  opposite  face  at  its  center  K  and  is  perpendic- 
ular to  this  face  also,  and  that  the  middle  point  0  of  KL  is 
the  point  required, 


- 

REGULAR  AND   SIMILAR  POLYHEDRONS.         101 

181.   PROBLEM.    To  inscribe  an  octahedron  in  a  sphere. 

Suggestion.  Making  use  of  the  con- 
struction, §  170  (3),  show  that  o  is  the 
point  equidistant  from  A,  B,  C,  D,  J0,  F. 

Give  all  the  steps  in  full. 

NOTE.  The  dodecahedron  and  icosahedron 
may  each  be  inscribed  in  a  sphere,  but  the 
proof  in  these  cases  is  much  more  complicated. 

182,  EXERCISES. 

1.  If  in  the  figure  of  §  176  the  tetrahedron  is  regular,  show  that 
OE  =  OF. 

2.  If  in  Ex.  1  a  sphere  is  described  with  center  0  and  radius  07?, 
would  it  touch  the  face  PBC  at  any  other  point  than  E't     (Why?) 
Would  any  part  of  the  surface  lie  on  the  opposite  side  of  ABC  from 
0  ?     (Why  V)     Is  the  same  true  of  each  of  the  other  faces  ? 

Such  a  sphere  is  said  to  be  inscribed  in  the  tetrahedron 
and  the  faces  are  said  to  be  tangent  to  the  sphere.  See 
§214. 

3.  In  the  figure  of  §  180  show  that  the  point  0  is  equidistant  from 
the  six  faces  of  the  cube  and  hence  that  a  sphere  may  be  inscribed. 

4.  In  the   figure  of   §  181  show  that  the  point  O  is  equidistant 
from  the  eight  faces  and  hence  that  a  sphere  may  be  inscribed. 

The  preceding  exercises  show  that  a  sphere  may  be 
inscribed  in  three  of  the  regular  polyhedrons,  and  the 
center  in  each  case  is  the  same  as  that  of  the  circumscribed 
sphere.  This  is  true  also  of  the  other  two  regular  poly- 
hedrons, but  the  proof  is  not  so  simple  as  in  these  cases. 
In  the  case  of  the  tetrahedron  a  sphere  may  be  inscribed 
whether  it  is  regular  or  not;  but  if  it  is  not  regular,  the 
center  is  not  the  same  as  that  of  the  circumscribed  sphere. 


102 


SOLID   GEOMETRY. 


SIMILAR  POLYHEDRONS. 

183,  Definitions.     Two  polyhedrons  are  similar  if  they 
have  the  same  number  of  faces  similar  each  to  each  and 
similarly  placed,  and  have  their  corresponding  polyhedral 
angles  congruent. 

Any  two  parts  which  are  similarly  placed  are  called 
corresponding  parts,  as  corresponding  faces,  edges,  vertices. 

184,  THEOREM.     Two    tetrahedrons   are  similar  if 
three  faces  of  one  are  similar  respectively  to  three  faces 
of  the  other,  and  are  similarly  placed. 


Given  the  tetrahedrons  P-ABC  and  P'-A'B'C1  having 
AAPB~  AA'PSf,  AAPC-AA'P1?,  and  A  BPC  ~  A  B'P'C. 

To  prove  p-ABC~Pf-AfBrcr. 

Proof:    (1)  Show  that  A  ABC~AAfB'cr. 
(2)  Show  that  trihedral  A  P  and  Pf  are  congruent. 
Likewise  Z  A  =*  Z  4',  Z  £  ^  Z  ur,  Z  C  ^  Z  c'. 
Hence,     by     definition     the 
polyhedrons  are  similar. 

185,  EXERCISES. 

1.  If  the  two  prisms  in  the  fig- 
ure are  similar,  name  the  pairs  of 
corresponding  parts.  Likewise  for 
two  similar  pyramids. 


REGULAR  AND  SIMILAR  POLYHEDRONS.         103 

2.  Show  that  a  plane  parallel  to  the  base  of  a  pyramid  cuts  off  a 
pyramid  similar  to  the  given  pyramid. 

SUGGESTION.  Use  the  principles  of  similar  triangles  and  §  66  to 
show  that  all  the  requirements  of  the  definition  (§  183)  are  fulfilled. 

3.  Does  a  plane  parallel  to  the  base  of  the  prism  cut  off  a  prism 
similar  to  the  given  prism  ?    Prove. 

4.  Show  that  two  tetrahedrons  are  similar  if  they  have  a  dihedral 
angle  in  one  equal  to  a  dihedral  angle  in  the  other  and  the  including 
faces  similar  each  to  each  and  similarly  placed. 

5.  Show  that  the  total  areas  of  two  similar  tetrahedrons  are  in  the 
same  ratio  as  the  squares  of  any  two  corresponding  edges. 

6.  Show  that  if  each  of  two  polyhedrons  is  similar  to  a  third  they 
are  similar  to  each  other. 

186.  THEOREM.  The  volumes  of  two  similar  tetra- 
hedrons are  in  the  same  ratio  as  the  cubes  of  their 
corresponding  edges. 

Given  P-ABC~  P'-A'&C',  with  volumes  Fand  V. 


To 


prove  that          —  -  =  .         • 
'          * 


P° 


Proof:   We  have     --  =    ,       '        '      ,  ,'    (§146) 
v'     P'A'  •  P'B'  •  P'C' 

Now  use  the  properties  of  similar  triangles  to  complete 
the  proof.     Use  the  figure  of  §  184. 

187,  EXERCISES. 

1.  Two  similar  tetrahedral  mounds  have  a  pair  of  corresponding 
dimensions  3  ft.  and  4  ft.     If  one  mound  contains  40  cu.  ft.  of  earth, 
how  much  does  the  other  contain  ? 

2.  The  edges  of  a  tetrahedron  are  3,  4,  5,  6,  7,  and  10.     Find  the 
edges  of  a  similar  tetrahedron  containing  64  times  the  volume. 

3.  Find  what  fraction  of  the  altitude  of  a  tetrahedron  must  be  cut 
off  by  a  plane  parallel  to  the  base,  measuring  from  the  vertex,  in 
order  that  the  new  pyramid  thus  detached  may  have  one  third  of  the 
original  volume. 


104 


SOLID   GEOMETRY. 


188,  Definitions.     Two  figures  are  said  to  have  a  center 
of  similitude  O,  if  for  any  two  points  A  and  B  of  the  one 
the  lines  AO  and  BO  meet  the  other  in  two  points,  A1  and 
Bf,  called  corresponding  points,  such  that 

AO  _  BO 

See  figures  under  §§  189-194. 

189,  THEOREM.    Any  two  figures  ivhich  have  a  center 
of  similitude  are  similar. 

Proof :    (1)   Two  triangles. 

Given  -04       OB       OG 


Let  the  student  prove  that  A  ABC~  A  A'B'C'. 
In  case  the  triangles  do  not  lie  in  the  same  plane,  use 
§  34  to  show  that  the  corresponding  A  are  equal. 

(2)   Two  polygons. 

Given  04       OBW    etc. 

l         m>l         nnt 


Give  the  proof  both  for  polygons  in  the  same  plane  and 
not  in  the  same  plane. 


' REGULAR  AND  SIMILAR  POLYHEDRONS.         105 
(3)   Two  tetrahedrons. 


With  the  same  hypothesis  as  before,  we  must  prove 
A  PAB~P'A'B',  A  PBC~A  P'B'C',  etc.,  and  then  use 
§184. 

(4)  Any  two  polyhedrons. 

(a)  Prove  corresponding  polygonal  faces  similar  to 
each  other. 

(£)  Prove  corresponding  polyhedral  angles  equal  to 
each  other. 

The  last  step  requires  not  only  equal  face  angles  about 
the  vertex,  as  in  the  case  of  the  tetrahedron,  but  also  equal 
dihedral  angles.  Note  that  two  dihedral  angles  are  equal 
if  their  faces  are  parallel  right  face  to  right  face  and  left 
face  to  left  face.  (Why?) 

(5)  Consider  any  two  figures  whatsoever  having  a  center 
of  similitude. 

(a)  Take  any  three  points  .4,  #,  C  in  one  figure  and 
the  three  corresponding  points  A',  Bf,  Cr  in  the  other. 

Then  AB  and  A'B',  AC  and  A'c',  etc.,  are  called  corre- 
sponding linear  dimensions,  and  the  triangles  ABC  and 
AfBrCf  are  corresponding  triangles. 

(6)  It  is  clear  that  any  two  corresponding  linear  dimen- 
sions have  the  same  ratio  as  any  other  two,  and  that  any 
two  corresponding  triangles  are  similar. 

In  this  sense  the  two  figures  are  said  to  be  similar. 


106  SOLID  GEOMETRY. 

190.  Definition.     The  ratio  of  similitude  of  two  similar 
figures  is  the  common  ratio  of  their  corresponding  linear 
dimensions.     This  ratio  is  the  same  as  the  distance  ratio 
of  corresponding  points  from  the  center  of  similitude. 

191.  THEOKEM.     Two  similar  triangles  may  be  so 
placed  as  to  have  a  center  of  similitude. 


A 
Given  the  similar  triangles  T  and  T,  in  which 


AB       AC       BC 

To  prove  that  they  may  be  placed  so  as  to  have  a  center 
of  similitude. 

Proof  :    From  any  point  O  draw  OA,  OJ5,  OC. 
On  these  rays  take  Av  Bv  C±  so  that 

OAl  _  OB1  _  OCj  _  ArBf 
OA~~  OB  ~~  OC~~   AB  ' 

Now  show  the  following  : 

(1)  A  T\  ~  A  T,  and  hence  A  TX  ~  A  Tr. 

(2)  AT^^AT'. 

For    this    show    that    A^  =  A'B'    by    means    of    the 


,.  ,I         ,        -, 

equations      11  =  —  1   and 


OA,        -,    AB      OA 


'' 


AB         OA  AB         OA 


REGULAR  AND  SIMILAR  POLYHEDRONS.         107 
Likewise  A1Cl  =  AtCr  and  BlCl=BrCf. 

(3)  Finally,  2^L—  °Ai 

OM       OA 

where  M  and  Ml  are  any  two  corresponding  points  what- 
ever. 

Hence  O  is  the  required  center  of  similitude. 

192,  THEOKEM.     Two   similar  tetrahedrons  may  lie 
so  placed  as  to  have  a  center  of  similitude. 

^^JZ 
p 

fk  /   121 


Given  the  similar  tetrahedrons  T  and  T. 

To  prove  that  they  can  be  placed  so  as  to  have  a  center 
of  similitude. 

Proof  :  With  O  as  a  center  of  similitude,  construct  Tv 

makinS  OA       OB  AB 

•  =  —  =  etc.  = 


OA,       OB, 


A'B1 


Now  show  as  in  §  191  that  T^  ^  T',  and  hence  that  Tr 
can  be  placed  in  the  position  TJ  so  as  to  have  with  T  the 
center  of  similitude  O. 

Give  all  the  steps  in  detail. 

193,  THEOREM.  Any  two  similar  polyhedrons  may 
be  placed  so  as  to  have  a  center  of  similitude. 

Suggestion.  The  argument  is  precisely  similar  to  that 
of  §  192.  Give  it  in  full. 


108 


SOLID   GEOMETRY. 


194,  In  the  figures  for  the  preceding  theorems  the  center 
of  similitude  has  been  taken  between  the  two  figures  or 
beyond  them  both.  The  center  may  be  taken  equally  well 
within  them,  as  in  the  following  illustrations  : 


D' 


B' 


In  the  case  of  similar  convex  polyhedrons  with  the 
center  of  similitude  thus  placed,  the  faces  are  the  bases  of 
pyramids  whose  vertices  are  all  at  the 
center  of  similitude. 

If,  further,  the  polygonal  faces  be 
divided  into  triangles  by  drawing  their 
diagonals,  these  triangles  become  the 
bases  of  tetrahedrons,  all  of  whose  ver- 
tices are  at  the  center  of  similitude. 

Moreover,  each  inner  tetrahedron  is 
similar  to  its  corresponding  outer  tetrahedron.  (Why?) 

The  volumes  of  the  two  similar  polyhedrons  are  thus 
composed  of  the  sums  of  sets  of  similar  tetrahedrons. 

195,  THEOREM.  The  volumes  of  any  two  similar 
polyhedrons  have  the  same  ratio  as  the  cubes  of  their 
corresponding  edges. 

Proof  :  Place  the  polyhedrons  whose  volumes  are  V  and 
V1  so  as  to  have  their  centers  of  similitude  within  them  as 
in  the  figures  of  §  194. 


REGULAR   AND   SIMILAR   POLYHEDRONS.          109 

Call  the  volumes  of  the  similar  tetrahedrons  Tv  T2,  T3,  •••, 
and  ay,  T2',  T3',  •••,  and  let  AB  and  AfBr  be  two  corre- 
sponding edges. 

Then  we  have 

Q 

AB    =  A  =  A  =  Ik  =  ....     (Why  ?) 

7 73  T>   I  T    I  rf,    I  ^  J         / 

A'  R'  ii  -/o  Jo 


And        ^^^-^^^      _=JLL  =  ::^_-.      (Why?) 

But  TJ  +  T2  +  T3  •••  =  Fand  T/  +  T2'  +  T3'  •••  =  Ff. 

F        Z§3 
Hence,  -7  =  ==r. 


196,  COROLLARY.  The  volumes  of  any  two  similar 
solids  are  in  the  same  ratio  as  the  cubes  of  any  two 
corresponding  linear  dimensions. 

This  proposition  may  be  rendered  evident  by  noticing  that  any  two 
similar  three-dimensional  figures  may  be  built  up  to  any  degree  of 
approximation  by  means  of  pairs  of  similar  tetrahedrons  similarly 
placed.  The  proposition  then  holds  of  any  two  corresponding  figures 
used  in  the  approximations. 

Note  that  the  ratio  of  similitude  of  two  similar  figures  may  be 
obtained  from  the  ratio  of  any  pair  of  corresponding  linear  dimensions. 

197.  EXERCISES. 

1.  If  two  coal  bins  are  of  the  same  shape  and  one  is  twice  as  long 
as  the  other,  what  is  the  ratio  of  their  cubical  contents  ? 

2.  What  is  the  ratio  of  the  lengths  of  the  two  bins  in  the  preced- 
ing example  if  one  holds  twice  as  much  coal  as  the  other  ? 

3.  Two  water  tanks  are  of  the  same  shape.     Find  the  ratio  of 
their  capacities  if  their  ratio  of  similitude  is  f . 

4.  In  the  preceding  what  must  be  the  ratio  of  similitude  in  order 
that  the  ratio  of  their  capacities  shall  be  f  ? 


110 


SOLID  GEOMETRY. 


APPLICATIONS  OF  SIMILARITY. 

198,  The  theorem  that  any  two  figures  which  have  a 
center  of  similitude  are  similar  is  the  geometric  basis  of 
many  mechanical  contrivances  for  enlarging  or  reducing 
both  plane  and  solid  figures ;  that  is,  for  constructing 
figures  similar  to  given  figures 
and  having  a  given  ratio  of 
similitude  with  them. 

The  essential  property  of  all 
such  contrivances  is  that  one 
point  O  is  kept  fixed,  while 
two  points  A  and  B  are  allowed 
to  move  so  that  O,  A,  and  B 
always  remain  in  a  straight 

line,  and  so  that  the  ratio   —  remains  the  same.     See 

OB 

§§  432-435  of  Plane  Geometry. 

In  the  first  figure  O  is  a  fixed  point.  Segments  OD,  CB, 
and  the  sides  of  the  parallelogram  AGED  are  of  fixed  length. 

Prove  that  if  B  is  once  so  taken  on  the  line  EC  as  to  be 
in  the  line  OA,  the  points  O,  -4,  and  B  will  always  remain 

collinear,  and  that  —  remains 
OB 

a  fixed  ratio. 

In  the  second  figure  is  shown 
an  ordinary  pantograph  used 
for  copying  and  at  the  same 
time  for  reducing  or  enlarg- 
ing maps,  designs,  etc.  The 
lengths  of  the  various  segments 
are  adjustable,  as  shown,  thus  obtaining  any  desired  scale. 

The  same  contrivance  may  be  used  for  copying  figures 
in  space  and  at  the  same  time  reducing  or  enlarging  them. 


REGULAR  AND   SIMILAR  POLYHEDRONS.         Ill 

199.  Now  consider  any  two  similar  figures  whatever  so 
placed  as  to  have  a  center  of  similitude  O.  We  have  seen 
that  if  points  AB  and  A'  B1  are  corresponding  points  of  the 
two  figures,  then  the  ratio  of  the  corresponding  linear 
dimensions  AB  and  A'B'  is  equal  to  the  ratio  of  similitude 


m 


-  of  the  two  figures. 


n 


Also  if  J,  B,  C,  D  and  A1 ',  #',  Cr,  Z>'  are  corresponding 
points,  then  A  ABC  and  ArBrCr,  and  the  tetrahedrons  ABCD 
and  A'B'C'D'  are  similar,  and  we  have 

Area  ^LBC        m2        -,     vol.  ABCD        m3 

-: — : — .  =  — r  and 


The  points  4,  .B,  C  and  A'B'C'  determine  two  planes,  each 
of  which  intercepts  a  certain  plane  figure  in  the  solid  figure 
to  which  the  points  belong.  These  two  plane  figures  we 
call  corresponding  cross  sections. 

We  assume  without  full  argument  that :  — 

THEOREM.  (1)  The  ratio  of  the  areas  of  any  pair 
of  corresponding  cross  sections  or  ajiy  pair  of  corre- 
sponding surfaces  of  similar  figures  is  equal  to  the 
square  of  their  ratio  of  similitude,  and 

(2)  The  ratio  of  the  volumes  of  any  two  similar  fig- 
ures is  equal  to  the  cube  of  their  ratio  of  similitude. 

Thus  the  ratio  of  the  radii  or  of  the  diameters  of  two  spheres  is 
their  ratio  of  similitude ;  likewise  the  ratio  of  the  lengths  or  of  the 
diameters  of  two  shells  used  in  gunnery,  or  the  ratio  of  the  heights  of 
two  men  of  similar  build. 

The  fact  that  the  ratio  of  the  areas  of  corresponding 
surfaces  of  similar  solids  is  equal  to  the  square  of  their 
ratio  of  similitude,  while  the  ratio  of  their  volumes  equals 
the  cube  of  this  ratio  is  one  of  the  most  important  and 
far-reaching  conclusions  of  geometry. 


112  SOLID   GEOMETRY. 


SUMMARY  OF   CHAPTER  IV. 

1.  Describe   the  five  regular  solids   according  to  the  form  and 
number  of  their  faces.     Why  can  there  not  be  more  than  these  five  ? 

2.  Compare  the  relation  of  the  regular  solids  to  the  sphere  with 
that  of  regular  polygons  in  the  plane  to  the  circle. 

3.  Review  the  process  of  construction  of  the  tetrahedron,  cube,  and 
octahedron. 

4.  Form   all  five   regular  polyhedrons  by  cardboard  models   as 
indicated  in  §  170. 

5.  Make  a  list  of  the  definitions  concerning  similar  polyhedrons. 

6.  Make  a  list  of  the  theorems  concerning  similar  polyhedrons. 

7.  Explain  the  relation  of  two  figures  which  have  a  center  of 
similitude. 

8.  What  theorem  of  this  chapter  is  referred  to  as  of  unusual  im- 
portance in  the  problems  and  applications? 

9.  State  the  applications  of  this  chapter  which  appeal  to  you  as 
especially  interesting  or  useful.     Return  to  this  question,  after  study- 
ing those  which  follow. 

PROBLEMS  AND  APPLICATIONS. 

1.  If  it  is  known  that  a  steel  wire  of  radius  r  will  carry  a  certain 
weight  w,  how  great  a  weight  will  a  wire  of  the  same  material  carry 
if  its  radius  is  2  r  ? 

SUGGESTION.     The  tensile  strengths  of  wires  are  in  the  same  ratio 
as  their  cross-section  areas. 

2.  Find  the  ratio  of  the  diameters  of  two  wires  of  the  same  mate- 
rial if  one  is  capable  of  carrying  twice  the  load  of  the  other ;  three 
times  the  load. 

3.  In  a  laboratory  experiment  a  heavy  iron  ball  is  suspended  by  a 
steel  wire.     In  suspending  another  ball  of  twice  the  diameter  a  wire 
of  twice  the  radius  of  the  first  one  is  used.     Is  this  perfectly  safe  if  it 
is  known  that  the  first  wire  will  just  safely  carry  the  ball  suspended 
from  it?     Discuss  fully. 

4.  In  two  schoolrooms  of  the  same  shape  (similar  figures)  but  of 
different  size,  the  same  proportion  of  the  floor  space  is  occupied  by 
desks.     Which  contains  the  larger  amount  of  air  for  each  pupil  ? 


REGULAR  AND  SIMILAR  POLYHEDRONS.          113 

5.  It  is  decided  to  erect  a  school  building  exactly  like  another 
already  built,  except  that  every  linear  dimension  is  to  be  increased 
by  ten  per  cent ;   that  is,  each  room  is  to  be  ten  per  cent   longer, 
wider,  and  higher,  and  so  for  all  parts  of  the  building.     If  the  air  in 
the  ventilating  flues  flows  with  the  same  velocity  in  the  two  build- 
ings, in  which  will  the  air  in  a  room  be  entirely  renewed  the  more 
quickly  ? 

SUGGESTION.  Note  that  the  ratio  of  the  amount  of  air  discharged 
by  two  flues  under  such  conditions  is  equal  to  the  ratio  of  their  cross- 
section  areas. 

6.  If  the  shells  used  in  guns  are  similar  in  shape,  find  the  ratio  of 
the  total  surface  areas  of  an  eight  inch  and  a  twelve  inch  shell. 

7.  Find  the  ratio  ,of  the  weights  of  the  shells  in  the  preceding 
problem,  weights  being  in  the  same  ratio  as  the  volumes. 

8.  If  a  man  5  ft.  9  in.  tall  weighs  165  lb.,  what  should  be  the 
weight  of  a  man   6  ft.  1  in.  tall,  supposing  them  to  be  similar  in 
shape  ? 

9.  What  is  the  diameter  of  a  gun  which  fires  a  shell  weighing 
twice  as  much  as  a  shell  fired  from  an  eight-inch  gun,  supposing  the 
shells  to  ba  similar  bodies? 

10.  The  ocean  liner  Mauretania  is  790  feet  in  length.     What  must 
be  the  length  of  a  ship  having  twice  her  tonnage,  supposing  the  boats 
to  be  similar  in  shape  ? 

11.  The  steamship  Lusitania  is  790  feet  long  with  a  tonnage  of 
32,500,  and  the  Olympic  is  882  feet  long  with  a  tonnage  of  45,000.     Are 
these  vessels  similar  in  shape?    If  not,  which  has  the  greater  capacity 
in  proportion  to  its  length  ? 

12.  Supposing  two  trees  to  be  similar  in  shape,  what  is  the  diame- 
ter of  a  tree  whose  volume  is  three  times  that  of  one  whose  diameter 
is  2  feet?    What  is  the  diameter  if  the  volume  is  5  times  that  of 
the  given  tree?    What  if  it  is  n  times  that  of  the  given  tree ? 

13.  Two  balloons  of  similar  shape  are  so  related  that  the  total 
surface  area  of   one  is  5  times  that  of  the  other.     Find  the  ratio  of 
their  volumes. 

See  page  166  for  further  applications. 


CHAPTER   V. 
THE  SPHERE. 

PLANE  SECTIONS   OF   THE  SPHERE. 

200,  Definitions.  A  sphere  consists  of  all  points  in  space 
which  are  equally  distant  from  a  fixed  point,  and  of  these 
points  only.  The  fixed  point  is  called  the  center  of  the 
sphere.  (See  §  177.) 

A  sphere  divides  space  into  two  parts  such  that  any 
point  which  does  not  lie  on  the  sphere  lies  within  it  or 
outside  it. 

The  sphere  may  be  developed  by  revolving  a  circle 
about  a  diameter  as  a  fixed  axis. 

A  line-segment  joining  any  two 
points  on  a  sphere  and  passing  through 
its  center  is  a  diameter.  A  segment 
joining  the  center  to  any  point  on  the 
sphere  is  a  radius. 

If  the  distance  from  a  point  to  the 
center  of  the  sphere  is  less  than  the 
radius,  the  point  is  within  the  sphere, 
and  if  greater  than  the  radius  it  is  out- 
side the  sphere. 

A  sphere  may  be  designated  by  a  single 
letter  at  its  center,  or  more  explicitly  by  naming 
its  center  and  radius. 

Thus  the  sphere  C  means  the  sphere  whose  center  is  C,  and  the 
sphere  CA  is  the  sphere  whose  center  is  C  and  whose  radius  is  CA . 

Two  spheres  are  said  to  be  equal  if  they  have  equal 
radii, 

114 


THE  SPHERE.  115 

201,  THEOREM.     All  radii  of  the  same  sphere  or  of 
equal  spheres  are  equal.     All  diameters  of  the  same 
sphere  or  of  equal  spheres  are  equal. 

These  statements  follow  directly  from  the  definitions. 

202,  EXERCISES. 

1.  How  does  the  definition  of  a  sphere  differ  from  that  of  a  circle? 
State  each  in  terms  of  a  locus. 

2.  If  two  spheres  have  the  same  center,  show  that  they  are  either 
equal  or  one  lies  entirely  inside  the  other. 

3.  In  how  many  points  can  a  straight  line  meet  a  sphere  ? 

4.  Does  every  line  through  an  interior  point  of  a  sphere  meet  it? 
In  how  many  points  ? 

203,  THEOREM.     A  section  of  a  sphere  made  ~by  a 
plane  is  a 


Given  a  sphere  with  center  C  cut  by  the  plane  M. 

To  prove  that  the  points  common  to  the  sphere  and  the 
plane  form  a  circle. 

Proof :  From  the  center  C  draw  CA  perpendicular  to 
the  plane  M. 

Let  B  and  D  be  any  two  points  common  to  the  plane 
and  the  sphere.  Complete  the  figure,  and  prove  AB  =  AD. 

Hence,  any  two  points  common  to  the  plane  and  the 
sphere  are  equidistant  from  A. 

How  must  this  proof  be  modified  in  case  the  plane  M 
passes  through  the  center  of  the  sphere  ? 


116 


SOLID    GEOMETRY. 


204,  Definitions.    A  circle  is  said  to  be  on  a  sphere  if  all 
its  points  lie  on  the  sphere. 

The  line  perpendicular  to  the  plane 
of  a  circle  at  its  center  is  called  the 
axis  of  the  circle. 

The  points  in  which  the  axis  of  a  ^ 
circle  on  a  sphere  meets  the  sphere  are 
called  the  poles  of  the  circle. 

If  the  plane  of  a  circle  on  a  sphere  passes  through  the 
centre  of  the  sphere,  it  is  called  a  great  circle  of  the  sphere, 
and  if  not,  it  is  called  a  small  circle. 

205,  THEOREM.     (1)   The   axis   of  any   circle  on  a 
sphere  passes  through  the  center  of  the  sphere. 


(2)  The  center  of  a  great  circle  is  the  center  of  the 
sphere. 

(3)  All  great  circles  are  equal  and  bisect  each  other. 

(4)  Three  points  on  a  sphere  determine  a  circle  on  the 
sphere. 

(5)  Through  two  given  points  on  a  sphere  there  is 
one  and  only  one  great  circle  unless  these  points  are  at 
opposite  ends  of  a  diameter. 

(6)  Every  great  circle  bisects  a  sphere. 

The  proofs  of  these  statements  follow  easily  from  the 
definitions.     Let  the  student  give  the  proofs  in  detail. 


THE  SPHERE. 


117 


206,  Definition.    The  distance  between  two  points  on  a 
sphere  is  the  distance  measured  between  these  points  along 
the  minor  arc  of  the  great  circle  through  them. 

207,  THEOREM.     All  points  of  a  circle  on  a  sphere 
are  equidistant  from  either  pole  of  the  circle. 


Given  P  a  pole  of  the  circle  whose  center  is  A,  and  let  B  and  D 
be  any  two  points  on  this  circle. 

To  prove  that  the  great  circle  arcs  PB  and  PD  are  equal. 
Suggestion.     Let  C  be  the  center  of  the  sphere. 
Prove  that  Z  ACE  =  Z  ACD. 
Hence,  chord  PB  =  chord  PD  and  PB  =  PD. 
Extend  the  radius  PC  to  meet  the  sphere  in  pf. 
Prove  that  P1  is  also  equidistant  from  any  two  points  of 
the  given  circle. 

208,  Definition.    The  common  distance  from  the  pole  of 
a  circle  to  all  points  on  it  is  called  the  polar  distance  of  the 
circle.     One  fourth  of  a  great  circle  is  a  quadrant. 

209,  COROLLARY  1.     The  polar  distance  of  a  great 
circle  is  a  quadrant. 

210,  COROLLARY  2.     If  a  point  p  is  at  a  quadrant's 
distance  from  each  of  two  points  not  at  the  extremities 
of  the  same  diameter,  it  is  the  pole  of  the  great  circle 
through  these  points. 


118  SOLID   GEOMETRY. 

211.  It  follows  from  the  preceding 
theorem  that,  if  a  spherical  blackboard 
is  at  hand,  circles  may  be  constructed 
on  it  by  means  of  crayon  and  string  the 
same  as  on  a  plane  blackboard.     Like- 
wise, curve-legged  compasses  may  be 
used. 

212,  EXERCISES. 

1.  How  many  small  circles  can  be  passed  through  two  points  on  a 
sphere?     How  many  great  circles ?     Show  why. 

2.  If  two  points  are  at  the  extremities  of  the  same  diameter  of  a 
sphere,  how  many  great  circles  can  be  passed  through  these  points  ? 

3.  What  great   circles  on  the  earth's  surface  pass  through  both 
poles?    If  a  great  circle  passes  through  one  pole,  must  it  pass  through 
the  other? 

4.  If  P  is  at  a  quadrant's  distance  from  each  of  two  points  A  and 
B,  and  if  these  points  are  at  opposite  ends  of  the  same  diameter,  is  P 
the  pole  of  any  circle  through  A  and  B(!    Of  how  many  circles? 

5.  If  A  and  B  are  at  opposite  ends  of  a  diameter,  can  a  small 
circle  be  passed  through  them  ? 

6.  If  two  circles  on  a  sphere  have  the  same  poles,  prove  that  their 
planes  are  parallel. 

7.  What  is  the  locus  of  all  points  on  a  sphere  at  a  quadrant's  dis- 
tance from  a  given  point  ? 

8.  What  is  the  locus  of  all  points  on  a  sphere  at  any  fixed  distance 
from  a  given  point  on  the  sphere  ?     What  is  the  greatest  such  dis- 
tance possible  ?     Discuss  fully. 

9.  If  two  planes  cutting  a  sphere  are  parallel,  compare  the  posi- 
tions of  the  poles  of  the  circles  thus  formed. 

10.  Find  the  locus  of  the  centers  of  a  set  of  circles  on  a  sphere 
formed  by  a  set  of  parallel  planes  cutting  it. 

11.  AB  is  a  fixed  diameter  of  a  sphere.     A  plane  containing  AB 
is  made  to  revolve  about  it  as  an  axis.    Find  the  locus  of  the  poles  of 
the  great  circles  on  the  sphere  made  by  this  revolving  plane.     How 
are  the  points  A  and  B  related  to  this  locus  ? 


THE  SPHERE.  119 

213,  THEOREM.  If  two  planes  cutting  a  sphere  are 
equidistant  from  the  center,  the  circles  thus  formed  are 
equal;  and  conversely, 

If  two  planes  cut  a  sphere  in  equal  circles,  the  planes 
are  equidistant  from  the  center. 


Suggestion.  In  the  figure  show  that  (1)  if  CA  =  CAr, 
then  AB  =AfBf,  and  (2)  if  AB  =  AfBr* 

^^0^B^^^. 

then  CA  =  CA' . 

214,  Definitions.      A  plane  which 
meets  a  sphere  in  only  one  point  is 
tangent  to  the  sphere. 

Two  spheres  are  tangent  to  each  other  if  they  have  only 
one  point  in  common. 

A  line  is  tangent  to  a  sphere  if  it  contains  one  and 
only  one  point  of  the  sphere. 

215.  EXERCISES. 

1.  If  a  plane  has  more  than  one  point  in  common  with  a  sphere, 
must  it  have  a  circle  in  common  with  the  sphere  ? 

2.  If  a  plane  is  tangent  to  a  sphere,  how  many  lines  in  the  plane 
are  tangent  to  the  sphere  ? 

3.  Can  two  spheres  be  tangent  to  each  other  and  still  one  be  in- 
side the  other  ? 


120 


SOLID   GEOMETRY. 


216.  THEOREM.  A  plane  tangent  to  a  sphere  is 
perpendicular  to  the  radius  from  the  point  of  tangency  ; 
and  conversely, 

A  plane  perpendicular  to  a  radius  at  its  extremity 
is  tangent  to  the  sphere. 


Given  a  sphere  C  with  plane  M  tangent  to  it  at  A. 

To  prove  that  CA  is  perpendicular  to  the  plane  M. 

Proof :  (1)  It  is  only  necessary  to  prove  that  CA  is  per- 
pendicular to  every  line  in  M  through  A.  (Why?) 

Draw  any  such  line  AB. 

The  plane  BAC  cuts  the  sphere  in  a  circle.  Prove  that 
AB  is  tangent  to  this  circle,  and  hence  perpendicular  to  AC. 

(2)  To  prove  the  converse,  note  that  CA  is  the  shortest 
distance  from  C  to  the  plane  M.  (Why?)  And  hence 
that  every  point  of  M  except  A  is  exterior  to  the  sphere. 

Hence,  M  is  a  tangent  plane.     (Why  ?) 

217,  Definition.     A  sphere  is  said  to  be  inscribed  in  a 
polyhedron  if  every  face  of  the  polyhedron  is  tangent  to 
the  sphere.    The  polyhedron  is  also  said  to  be  circumscribed 
about  the  sphere. 

218,  THEOREM.     A  sphere  may  l)e  inscribed  in  any 
tetrahedron. 

The  proof  is  left  to  the  student.     See  §  182. 


THE  SPHERE.  121 

219.  EXERCISES. 

1.  How  many  planes  may  be  tangent  to  a  sphere  at  one  point  on 
the  sphere  ?     How  many  lines  ? 

2.  Through  a  given  point  exterior  to  a  sphere  construct  a  line 
tangent  to  the  sphere. 

SUGGESTION.  Let  0  be  the  center  of  the  sphere  and  P  the  given 
exterior  point.  Pass  any  plane  M  through  P  and  0.  In  the  plane 
M  construct  a  line  through  P  tangent  to  the  great  circle  in  which  M 
cuts  the  sphere. 

3.  How  many  lines  tangent  to  a  sphere  can  be  constructed  from  a 
point  outside  the  sphere? 

4.  Through  a  given  point  exterior  to  a  sphere  construct  a  plane 
tangent  to  the  sphere. 

SUGGESTION.  As  in  Ex.  2  construct  a  line  through  the  given  point 
tangent  to  the  sphere.  Through  the  point  of  tangency  of  this  line 
pass  a  plane  tangent  to  the  sphere. 

5.  How  many  planes  can  be  passed  through  a  given  exterior  point 
tangent  to  the  sphere  ? 

6.  How  many  planes  tangent  to  a  sphere  can  be  passed  through 
two  given  points  A  and  B  outside  a  sphere?     Discuss  fully  if  the  line 
AB  (1)  meets  the  sphere  in  two  points;  (2)  is  tangent  to  the  sphere  ; 
(3)  does  not  meet  the  sphere. 

220,  THEOREM.  The  intersection  of  two  spheres  is 
a  circle. 

Proof:  The  two  intersecting 
spheres  may  be  developed  by 
rotating  about  a  fixed  axis  CCf 
two  intersecting  circles  with 
centers  C  and  C1 . 

Let  A  and  B  be  the  two  points  common  to  both  circles. 
Then  BA  -L  ccr.  (Why  ?) 

As  the  figure  rotates  about  the  line  CCr,  AB  remains 
fixed  in  length  and  perpendicular  to  CCf. 

Hence,  B  traces  out  a  circle.     See  §  23. 


122 


SOLID   GEOMETRY. 


221,  PKOBLEM.  To  find  the  diameter  of  a  given 
material  sphere. 

With  any  point  P  of  the  sphere  as  a  pole,  construct  any 
circle,  and  on  this  circle  select  any  three  points  A,  B,  C. 

Using  a  pair  of  compasses,  measure  the  straight  line- 
segments  AB,  BC,  (Li,  and  construct  the  triangle  AfBrCf 
congruent  to  ABC. 

Let  BfDf  be  the  radius  of  the  circle  circumscribed  about 
A'B'C'. 

If  ppr  is  the  axis  of  the  circle  ABC  on  the  sphere  and 
BD  the  radius  of  this  circle,  then  BD  =  BrDr. 


Measure  PB  by  means  of  the  compasses. 

Then  P.BP'  is  a  right  triangle,  with  BD  perpendicular 
to  its  hypotenuse  PP'. 

PB  and  BD  being  known,  we  may  now  compute  PD  from 
the  right  triangle  PBD  and  then  compute  PP'  from  the 
similar  triangles  PSD  and  PP'D,  finding  PD:  PB  —  PB  :  PP1 
or  PD  x  PP'  =  PB2.  See  Ex.  3,  §  66,  Plane  Geometry. 

The  segment  PP'  may  also  be  found  by  geometric  con- 
struction ;  namely,  draw  a  triangle  congruent  to  P1  BP. 

Show  how  to  do  this  when  BP  and  BD  are  known. 


THE  SPHERE.  123 

222,  EXERCISES. 

1.  How  many  points  are  necessary  to  determine  a  sphere  ?     See 
§177. 

2.  If  the  center  of  a  sphere  is  given,  how  many  points  on  the 
sphere  are  required  to  determine  it? 

3.  If  a  plane  M  is  tangent  to  a  sphere  at  a  point  A ,  show  that 
the  plane  of  every  great  circle  of  the  sphere  through  A  is  perpendicular 
toM. 

4.  Show  that  the  line  of  centers  of  two  intersecting  spheres  meets 
the  spheres  in  the  poles  of  their  common  circle. 

5.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  a  given 
plane  at  a  given  point. 

6.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  a  given 
line  at  a  given  point. 

7.  Find  the  locus  of  the  centers  of  all  spheres  of  given  radius 
tangent  to  a  fixed  plane. 

8.  Find  the  locus  of  the  centers  of  all  spheres  of  given  radius 
tangent  to  a  fixed  line. 

9.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  two 
given  intersecting  planes. 

10.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  all  faces 
of  a  trihedral  angle. 

11.  Show  that  two  spheres  are  tangent  if  they  meet  on  their  line 
of  centers.     Distinguish  two  cases.     Compare  §  209,  Plane  Geometry. 

12.  State  and  prove  the  converse  of  the  preceding  proposition. 

13.  In  plane  geometry  how  many  circles  can  be  drawn  through  a 
given  point  tangent  to  a  given  line  at  a  given  point  ? 

14.  If  a  given  sphere  is  tangent  to  a  given  plane  M  at  a  given 
point  A,  how  many  points  on  the  sphere  are  required  to  determine  it? 

SUGGESTION.     Suppose  one  point  P  given.     Pass  a  plane  through 
P  _L  to  plane  M  at  A .     Is  there  only  one  such  plane?     Discuss  fully. 

15.  Describe  the  set  of  all  lines  in  space  whose  distances  from  the 
center  of  a  sphere  are  all  equal  to  the  radius  of  the  sphere. 

16.  Describe  the  set  of  all  planes  whose  distances  from  the  center 
of  a  sphere  are  all  equal  to  the  radius  of  the  sphere. 


124  SOLID   GEOMETRY. 


TRIHEDRAL  ANGLES  AND   SPHERICAL   TRIANGLES. 

223,  Definitions.  When  two  curves  meet,  they  are  said 
to  form  an  angle ;  namely,  the  angle  made  by  the  tangents 
to  the  curves  at  their  common  point. 

Any  two  planes  through  the  center  of  a  sphere  cut  out 
two  great  circles  which  intersect  in  two  points  and  form 
four  spherical  angles  about  each  of  these  points.  Two  of 
these  angles  with  a  common  vertex  are  either  adjacent  or 
vertical  in  the  same  sense  as  the  angles  formed  by  two  in- 
tersecting straight  lines. 

A  spherical  angle  is  acute,  right,  or  obtuse  according  to 
the  form  of  the  angle  between  the  tangents  to  its  sides 
(arcs)  at  their  common  point. 

Any  two  circles  on  a  sphere  which  meet,  whether  great 
circles  or  not,  form  angles  according  to  the  above  defini- 
tion. 

Only  angles  formed  by  great  circles  are  considered  in 
this  book  and  the  expression  spherical  angle  will  be  under- 
stood to  refer  to  such  angles.  Only  angles  greater  than 
zero  and  not  greater  than  two  right  angles  are  considered. 

A  spherical  angle  may  be  denoted  by  a  single  letter  or  by  three 
letters,  as  in  the  case  of  a  plane  angle. 

224,  THEOREM.  A.  spherical  angle 
is  measured  ~by  an  arc  of  the  great 
circle  whose  pole  is  the  vertex  of  the 
angle  and  which  is  intercepted  by 
the  sides  of  the  angle. 

Suggestion.     Show   that   AB   meas- 
ures the  dihedral  angle  formed  by  the  planes  PAG  and 
PEG  and  that  /.BCA  = 


THE  SPHERE.  125 

225.  Definitions.     The  section  of  a  sphere  made  by  a 
convex  trihedral  angle,  whose  vertex  is  at  the  center  of 
the  sphere,  is  called  a  spherical  triangle. 

The  face  angles  of  the  trihedral  angle 
are  measured  by  the  sides  (arcs)  of  the 
spherical  triangle,  and  its  dihedral 
angles  are  equal  to  the  angles  of-  the 
spherical  triangle. 

Since  each  face  angle  of  a  trihedral  angle  is  less  than  two  right 
angles,  it  follows  that  each  side  of  a  spherical  triangle  is  less  than 
a  semicircle. 

226,  EXERCISES. 

1.  Show  that  a  spherical  angle  is  equal  to  the  plane  angle  of  the 
dihedral  angle  formed  by  the  planes  of  the  great  circles  whose  arcs 
are  the  sides  of  the  spherical  angle.     See  figure  under  §  224. 

2.  Prove  that  vertical  spherical  angles  are  equal. 

3.  Prove  that  the  sum  of  the  spherical  angles  about  a  point  is  four 
right  angles. 

4.  At  what  angle  does  a  meridian  on  the  earth's  surface  intersect 
the  equator  ? 

5.  Denote  by  P  the  North  Pole  on  the  earth's  surface.     Consider 
any  two  meridians  forming  an  angle  of  one  degree  at  P  and  meeting 
the  equator  in  points  A  and  B,  respectively.     What  is  the  sum  of  the 
angles  of  the  spherical  triangle  PAB1!    Compare  with  the  sum  of 
the  angles  of  »a  plane  triangle. 

6.  Is  it  possible  to  construct  a  spherical  triangle  each  of  whose 
angles  is  a  right  angle? 

SUGGESTION.  Consider  twb  meridians  forming  a  right  angle  at  P. 
Such  a  triangle  is  called  a  trirectangular  triangle. 

7.  In  a  trirectangular  spherical  triangle  what  is  the  length  qf  each 
side  in  terms  of  degrees  ? 

The  question  as  to  whether  or  not  the  sum  of  the  angles  of  a 
spherical  triangle  is  ever  equal  to  two  right  angles  is  answered  in 
§257. 


126 


SOLID   GEOMETRY. 


227,    THEOREM.     TJie  sum  of  two  face  angles  of  a 
trihedral  angle  is  greater  than  the  third  face  angle. 

c 


Proof :  Connect  points  A  and  D  on  two  sides. 

Suppose  not  all  three  face  angles  are  equal  and  that 
Z.  ACD  >  /.  AGE. 

Construct  CE  in  the  face  ACD,  making  Z.  ACE  =  £  ACB. 

Lay  off  CB  =  CE  and  draw  AB  and  BD. 

Now  show  that  (1)  AB  =  AE,  (2)  AD  <  AB  +  BD, 
(3)  ED  <  BD,  (4)  Z.  ECD  <  £  BCD. 

Hence,  /.  ACD  <  Z.  ACB  -f-  Z.  BCD. 

228,  COROLLARY  1.     The  sum  of  two  sides  of  a  spheri- 
cal triangle  is  greater  than  the  third  side. 

229,  COROLLARY  2.     The  sum  of  the  three  sides  of  a 
spherical  triangle  is  less  than  a  great  circle. , 

Proof:  AC Ar  +ABA'=  a  great  circle. 
But  AC  <  ACAr  and  AB  <  ABA1,  ' 
and  'BC  <  BA'  +  CA'  by  Corollary  1. 
Hence,  AC  +  CB  +  BA  <  a  great  circle. 

230,  COROLLARY  3.     State  and  prove  the  theorems  on 
trihedral  angles  corresponding  to  Corollaries  1  and  2. 


THE  SPHERE. 


127 


231,  THEOREM.     The  shortest  distance  on  a  sphere  'be- 
tween two  of  its  points  is  measured  along  the  minor 
arc  of  a  great  circle  passing  through  these  points. 

Proof:  Let  A  and  B  be  any  two 
points  on  a  sphere,  AB  the  minor  arc 
of  a  great  circle  through  them,  and 
ADCB  any  other  curve  on  the  sphere 
connecting  A  and  B  with  points  C  and 
D  on  it  in  the  order  ADCB.  Neither 
(7  nor  D  is  on  the  arc  AB. 

Draw  the  great  circle  arcs  AD,  AC,  DC,  and  CB.     Then 

by  §  228  AC  -f-  CB  >  AB  and  AD  +  DC  >  AC. 

Hence,  AD  -f-  DC  +  CB  >  AB. 

Continuing  in  this  manner,  we  obtain  a  succession  of 
paths,  each  longer  than  the  preceding.  But  by  this  process 
we  get  closer  and  closer  to  the  length  of  the  curve  ACB. 

Hence,  it  must  be  greater  than  that  of  AB. 

232,  Definitions.    Two  trihedral  angles  are  symmetrical 
one  to  the  other  if  the  face  angles  and  the  dihedral  angles 
of  one  are  equal  respectively  to  the  face  angles  and  the 
dihedral    angles 

of  the  other,  but 
arranged  in  the 
opposite  order. 

Similarly,  two 
spherical  trian- 
gles are  sym- 
metrical one  to 
the  other  if  the  sides  and  the  angles  of  one  are  equal 
respectively  to  the  sides  and  the  angles  of  the  other,  but 
arranged  in  the  opposite  order. 


128  SOLID  GEOMETRY. 

233,  THEOKEM.  If  the  radii  drawn  from  the  ver- 
tices of  a  spherical  triangle  are  extended,  they  meet 
the  sphere  in  the  vertices  of  a  triangle  symmetrical  to 
the  given  triangle. 


Ar 

The  proof  is  left  to  the  student. 

234,  COROLLARY.     State  and  prove  the  correspond- 
ing theorem  on  trihedral  angles. 

235,  THEOREM.     Tivo  trihedral  angles  having  their 
vertices  at  the  center  of  the  same  or  of  equal  spheres 
intercept  congruent  spherical  triangles  if  the  trihedral 
angles   are  congruent.,  and  symmetrical  spherical   tri- 
angles if  they  are  symmetrical. 


This  is  an  immediate  consequence  of  §§  225,  232. 

236,  COROLLARY.  If  in  two  spherical  triangles  three 
sides  of  one  are  equal  respectively  to  three  sides  of  the 
other,  and  arranged  in  the  same  order,  the  triangles  are 
congruent.  See  §  66. 


4  *-> 

THE  SPHEBE.  129 

237,  THEOREM.     If  the  face  angles  of  one  trihedral 
angle  are  equal  respectively  to  the  face  angles  of  another, 
but  arranged  in  the  opposite  order,  the  trihedral  angles 
are  symmetrical. 

Proof:  By  §  64  the  dihedral  angles  of  one  are  equal 
respectively  to  those  of  the  other.  Now  verify  that  these 
parts  are  arranged  in  the  opposite  order. 

238,  COROLLARY  1.    If  in  two  spherical  triangles  the 
three  sides  of  one  are  equal  to  three  sides  of  the  other, 
~but  arranged  in  the  opposite  order,  the  triangles  are 
symmetrical. 

239,  Definition.     A  spherical  triangle  is  isosceles  if  two 
sides  are  equal. 

240,  COROLLARY  2.     The  angles 
opposite  the  equal  sides  of  an  isos- 
celes spherical  triangle,  are  equal. 

Suggestion.     Let  AC  and  BC  be  the  A* 
equal  sides.     Draw  CD  to  the  middle 
point  of  AB. 

241,  COROLLARY  3.     If  two  isosceles  spherical  tri- 
angles are  symmetrical,  they  are  congruent,  and  con- 
versely. 

242,  THEOREM.     If  two  trihedral  angles  are  sym- 
metrical to  the  same  trihedral  angle,  they  are  congruent. 

Suggestion.  Show  that  the  corresponding  parts  must 
be  arranged  in  the  'same  order. 

243,  COROLLARY.     State  and  prove  the  correspond- 
ing theorem  for  spherical  triangles. 


130 


SOLID   GEOMETRY. 


244,  THEOREM.  Two  spherical  triangles  having  two 
sides  and  the  included  angle  of  one  equal  respectively  to 
two  sides  and  the  included  angle  of  the  other  are  con- 
gruent, if  the  given  parts  are  arranged  in  the  same 
order,  and  symmetrical,  if  they  are  arranged  in  the  op- 
posite order. 
A 


Proof:  If  the  given  parts  are  arranged  in  the  same 
order,  the  proof  may  be  made  by  superposition  exactly  as 
in  §  32,  Plane  Geometry. 

If  the  given  parts  are  arranged  in  the  opposite  order, 
proceed  as  follows : 

Denote  the  given  triangles  by  t±  and  £2.  Construct  a 
spherical  triangle  £3  symmetrical  to  tr  Then  by  §  243 
£2  and  £3  are  congruent.  Hence,  if  t±  is  symmetrical  to  £3, 
it  must  be  symmetrical  to  £2. 

245,  COROLLARY.  State  and  prove  the  correspond- 
ing theorem  for  trihedral  angles. 


246, 


EXERCISES. 


1.  Compare  fully  the  theorems  on  the  congruence  of  plane  tri- 
angles and  of  trihedral  angles.     Is  there  any  theorem  in  either  case 
for  which  there  is  no  corresponding  theorem  in  the  other? 

2.  Compare  in  the  same  manner  the  theorems  on  the  congruence 
of  plane  triangles  and  of  spherical  triangles. 

3.  Compare  in  the  same  manner  the  theorems  on  the  congruence 
of  trihedral  angles  and  of  spherical  triangles. 


THE  SPHERE.  131 

247.  THEOREM.  The  locus  of  all  points  on  a  sphere 
equidistant  from  two  fixed  points  on  the  sphere  is  a 
great  circle  bisecting  at  right  angles  the  great  circle  arc 
connecting  the  two  given  points. 


Proof:    Let  C  be  the  middle  point  of  AB. 

(a)  If  AP  =  BP  prove  that  A  ACP  and  BCP  are  sym- 
metrical, arid  hence,  Z  ACP  =  Z  BCP  =  rt.  Z. 

(6)  If  Z^lCP  =  Z  BCP,  prove  that  AP  =  BP. 
Why  are  steps  (V)  and  (7>)  both  needed  ? 

248,  EXERCISES. 

1.  If  two  face  angles  of  a  trihedral  angle  are  equal,  the  opposite 
dihedral  angles  are  equal. 

2.  If  two  face  angles  of  a  trihedral  angle  are  equal,  it  is  congruent 
to  its  symmetrical  trihedral  angle. 

3.  Show  how  to  find  a  pole  of  the  circle  through  three  given  points 
on  the  sphere. 

SUGGESTION.  Let  the  given  points  be  A,  B,  C.  By  §  207  the 
pole  of  the  circle  is  equidistant  from  A,  B,  and  C.  Connect  A 
and  B  by  an  arc  of  a  great  circle  and  construct  another  arc  of  a  great 
circle  bisecting  AB  perpendicularly.  Similarly  construct  a  perpen- 
dicular bisector  of  BC .  The  points  in  which  these  two  arcs  meet  will 
be  the  poles  of  the  circle  through  A ,  B,  and  C. 

State  this  argument  in  full  detail. 


132 


SOLID   GEOMETRY. 


POLAR  TRIANGLES. 

249,  Definition.  With  the  ver- 
tices A,  B,  C  of  a  spherical  tri- 
angle as  poles,  construct  three 
great  circles.  Each  of  these 
^circles  meets  each  of  the  others 
in  two  points,  thus  forming  eight 
spherical  triangles,  as  shown 
in  the  figure,  namely,  A'B'C', 
ArsrF,  B'C'D,  C'A'E,  A'EF,  B'DF, 
C'*DE,  and  DEF. 

There  is  one  and  only  one  of  these,  namely,  ArBrCf,  such 
that  A  and  A'  are  on  the  same  side  of 
.B^C?,  B  and  B'  on  the  same  side  of  A*~c', 
and    C  and    c'   on   the   same   side    of 
A'B'. 

The    triangle    ArBfCr   as    thus   de- 
scribed is  the  polaf  triangle  of  ABC. 


250, 


EXERCISES. 


1.  In  the  figure  the  parts  of  the  great  circles  which  are  supposed 
to  be  on  the  front  side  of  the  figure  are  given  in  solid  lines  while  the 
parts  on  the  back  side  are  dotted.     Study  the  figure  with  care  and 
state  which  triangles  are  entirely  on  the  front  side,  which  are  entirely 
on  the  back  side  of  the  sphere,  and  which  are  partly  on  the  front  side 
and  partly  on  the  rear  side  of  the  sphere. 

2.  Show  that  the  points  A'  and  D  cannot  be  on  the  same  side  of 
the  circle  through  B'C'. 

SUGGESTION.     Can  the  two  extremities  of  a  diameter  lie  in  the 
same  hemisphere  ? 

3.  If  A  is  a  pole  of  the  great  circle  through  B'C'  and  if  A'  is  on 
the  same  side  of  this  circle  as  A,  show  that  A  and  A'  are  less  than 
one  quadrant's  distance  apart. 


v/y 

THE  SPHERE.  133 

251,  THEOREM.     If  A'B'C'  is  the  polar  triangle  of 
ABC,  then  ABC  is  the  polar  triangle  of  A'B'C'. 

Proof:  It  is  required  to  prove  (1)  that  A1  is  the  pole  of 
JIG,  B'  the  pole  of  AC,  and  C1  the  pole  of  AB,  and  also  (2) 
that  A  and  Af  lie  on  the  same  side  of  BC,  B  and  Bf  on 
the  same  side  of  AC,  and  C  and  c1  on  the  same  side  of  AB. 

(1)  To  prove  that  Ar  is  the  pole  of  J3C  we  need  only  to 
show  that  A1  is  at  a  quadrant's  distance  from  two  points 
in  BC.     Why  ? 

Now  A'  is  at  a  quadrant's  distance  from  B  because  B  is 
the  pole  of  A'C'.  A1  is  also  at  a  quadrant's  distance  from 
C  because  C  is  the  pole  of  A'B'.  Hence,  Ar  is  the  pole  of 
BC. 

Similarly,  Bf  is  the  pole  of  AC  and  c'  the  pole  of  AB. 

(2)  To  show  that  A  and  A'  lie  on  the  same  side  of  the 
circle  BC,  we  note  that  since  A  is  the  pole  of  the  circle  BfCf 
and  A  lies  on  the  same  side  of  this  circle  with  Ar,  then  A 
and  A1  are  at  less  than  a  quadrant's  distance.     Hence,  it 
follows  that  if  A1  is  at  a  quadrant's  distance  from  BC,  A 
and  Af  must  be  on  the  same  side  of  BC. 

In  like  manner  we  show  that  B  and  Bf  lie  on  the  same 
side  of  AC  and  C  and  cf  on  the  same  side  of  AB. 

252,  Definition.    If  ABC  and  ArBfCr  are  polar  triangles, 
and  if  A  is  a  pole  of  fc*,  then  /.A  and  B'C'  are  said  to  be 
corresponding  parts. 

253,  EXERCISES. 

1.  In  the  two  polar  triangles  ABC  and  A'B'C'  name  all  pairs  of 
corresponding  parts. 

2.  Is  there  any  spherical  triangle  such  that  its  polar  triangle  is 
identical  with  the  given  triangle? 

3.  If  one  side  of  a  spherical  triangle  is  greater  than  a  quadrant  show 
that  it  is  cut  by  two  circles  in  the  construction  of  its  polar  triangle. 


134 


SOLID   GEOMETRY. 


254,  THEOREM.  The  sum  of  the  measures  of  an 
angle  of  a  spherical  triangle  and  the  corresponding  arc 
of  its  polar  triangle  is  180°. 

Given  the  polar  triangles  ABC  and  A'B'C'.  Denote  the  meas- 
ures in  degrees  of  the  angles  by  A}  B,  C,  •••  and  of  the  correspond- 
ing sides  by  a',  b',  c',  —.  A. 


a  D 

To  prove  that       A  +  af  =  180° 


Proof:  Extend  (if  necessary)  arcs  A1  B'  and  A'c'  till  they 
meet  the  great  circle  BC  in  points  D  and  E,  respectively. 
Then  arc  DE  is  the  measure  of  Z  Ar. 

Also  SS  =JW°,  and  5c=90°.     (Why?) 

But  BE  +  DC  =  BC  +  ED  =  a  +  Ar. 

Hence,  A1  +  a  =  180°. 

Complete  the  proof  for  the  other  cases. 

255,  COROLLARY.  If  two  spherical  triangles  are  con- 
gruent or  symmetrical,  their  polar  triangles  are  con- 
gruent or  symmetrical. 


256, 


EXERCISES. 


1.  Does  the  above  proof  apply  to  the  second  figure  ? 

2.  If  two  angles  of  a  spherical  triangle  are  equal,  it  is  isosceles. 
SUGGESTION.     Use  §  254,  §  240,  and  again  §  254. 


THE  SPHERE.  1J5 

3.  State  and  prove  the  theorem  on  trihedral  angles  corresponding 
to  the  preceding.  c 

4.  If  two  angles  of  a  spherical  triangle  are 
unequal,  the  sides  opposite  them  are  unequal, 
the  greater  side  being  opposite  the  greater 
angle. 

SUGGESTION.  In  the  triangle  ABC  let  ZB>/.A.  Draw  Bl), 
making  Z.AED  =  ZA. 

Then,  AD  =  BD,  and  BD  +  DC  >  BC. 

Hence,  show  that  AC  >  BC. 

5.  State  and  prove  a  theorem  on  trihedral  angles  corresponding  to 
the  preceding. 

257,  THEOREM.  The  sum  of  the  angles  of  a  spheri- 
cal triangle  is  less  than  six  right  angles  and  greater  than 
two  right  angles. 


B    — _ 

a 

Given  the  spherical  triangle  ABC. 
To  prove  that  (1)  Z^L4-Z£-r-Zc<6rt.  angles. 
(2)  Z  A  +  Z-  B  +  Z.  C>  2  rt.  angles. 
Proof:  Construct  the  polar  triangle  AfBfCf,  with  sides 
a',  ft',  c'. 

(1)  By  §  254  Z  .4  +Z  B+Z  C+a'  +£'  +  <?'=  6  rt.  angles. 
Since  af  -f  b1  +  c1   is  greater  than  zero,  it  follows  that 

Z^+ZjB  +  Zc<6rt.  angles. 

(2)  Using  §  229,  show  that  /.A  +  Z  B  +  Z  c>  2  rt.  A. 

258,  COROLLARY.     State  and  prove  the  theorem  on 
trihedral  angles  which  corresponds  to  the  preceding. 


136 


SOLID   GEOMETBY. 


259,   PROBLEM.     On  a  given  sphere  to  construct  a 
spherical  triangle  when  its  sides  are  given. 


D, 


Solution.  Let  o  be  the  given  sphere,  and  #,  5,  c  the  arcs 
of  the  required  triangle,  and  let  A  A'  be  any  diameter  of 
the  sphere.  With  A  as  a  pole,  construct  circles  DBE  and 
FCG  whose  polar  distances  from  A  are  c  and  b  respectively. 

With  B  as  a  pole,  construct  a  circle  HCK,  whose  polar 
distance  from  B  is  a.  Then  construct  the  three  great 
circle  arcs,  AB,  J3C,  CA.  ABC  is  the  required  triangle. 


260, 


EXERCISES. 


1.  What  restrictions  if  any  is  it  necessary  to  impose  upon  the  three 
given  sides  of  the  triangle  in  §  259  V     (§§  228,  229.) 

2.  In  plane  geometry  two  congruent  triangles  may  be  constructed 
upon  the  same  base  and  on  the  same  side  of  it.     Is  a  corresponding 
construction  possible  on  the  sphere  ? 

3.  If  in  the  above  construction  each  of  two  sides  of  the  required 
triangle  is  very  great,  that  is,  nearly  a  semicircle,  show  from  the  con- 
struction that  the  third  side  must  be  very  small. 

4.  If  one  side  of  the  proposed  triangle  in  §  259  were  equal  to  or 
greater  than  180°,  why  would  that  make  the  construction  impossible  ? 


y/s- 

THE  SPHEBE.  137 

261,  PROBLEM.      To  construct  a  spherical  triangle 
when  its  three  angles  are  given. 

Solution.  Let  the  three  given  angles  be  A,  .B,  C,  and  let 
a',  V,  c'  be  arcs  such  that  a1  +  ^  A  =  180°,  br  +  Z  E  =  180°, 
c'  +  Z  c  =  180°.  Then  the  triangle  whose  arcs  are  a',  &',  c' 
will  be  the  polar  triangle  of  the  required  triangle.  This 
latter  triangle  A'B'C'  may  be  constructed  by  the  method 
of  §  259.  Then  construct  the  polar  triangle  of  A'B'C', 
which  will  be  the  required  triangle. 

Give  reasons  in  full  for  each  step. 

262,  PROBLEM.     To  construct  a  trihedral  angle  when 
its  face  angles  are  given. 

Solution.  Construct  the  corresponding  spherical  triangle 
by  the  method  of  §  259. 

Give  the  construction  in  full  and  prove  each  step. 

263,  PROBLEM.     To  construct  a  trihedral  angle  when 
its  dihedral  angles  are  given. 

Solution.     Construct  the  corresponding   spherical   tri- 
angle by  the  method  of  §  261. 
Give  reasons  in  full  for  each  step. 

264,  EXERCISES. 

1.  If  two  spherical  triangles  having  angles  respectively  equal  are 
constructed  on  the  same  sphere,  how  are  these  triangles  related?  Prove. 

2.  If  two  trihedral  angles  with  face  angles  respectively  equal  are 
constructed  as  in  §  262,  how  are  they  related  ?    Prove. 

3.  If  two  trihedral  angles  each  with  the  same  dihedral  angles  are 
constructed  as  in  §263,  how  are  the  trihedral  angles  related?    Prove. 

4.  What  restrictions  if  any  must  be  placed  upon  the  given  angles 
A,B,  C  in  §  261  ?     Compare  Ex.  1,  §  260. 

5.  What  restrictions  if  any  are  needed  in  Exs.  2  and  3  ? 


138  SOLID   GEOMETRY. 

265,  THEOREM.     Two  spherical  triangles  having  two 
angles  and  the  included  side  of  one  equal  respectively  to 
two  angles  and  the  included  side  of  the  other  are  con- 
gruent, if  the  given  equal  parts  are  arranged  in  the 
same  order,  and  symmetrical,  if  arranged  in  the  oppo- 
site order. 

Proof:  By  §§254  and  244  the  polar  triangles  of  the  given 
triangles  are  congruent  or  symmetrical.  Hence,  by  §  255, 
the  given  triangles  themselves  are  congruent  or  symmet- 
rical. 

266,  COROLLARY.     State  and  prove  the  theorem  on 
trihedral  angles  which  corresponds  to  the  preceding. 

267,  THEOREM.     Two  spherical  triangles  having  the 
angles  of  one  equal  respectively  to  the  angles  of  the 
other  are  congruent,  if  the  equal  angles 

are  arranged  in  the  same  order,  and 
symmetrical,  if  they  are  arranged  in 
the  opposite  order. 

Proof:    By   §§254,   236,    and  238  the 
polar    triangles    of    the    given    triangles 
are    equal    or  symmetrical.     Hence,    by 
§  255,  the  given  triangles  themselves  are  equal  or  sym- 
metrical. 

268,  COROLLARY.     State  and  prove  the  theorem  on 
trihedral  angles  ivhich  corresponds  to  the  preceding. 

269,  Is  there  a  theorem  on  plane  triangles  correspond- 
ing to  that  of  §  267  ? 


• 

THE  SPHERE.  139 


270,  EXERCISES. 

1.  If  the  sides  of  a  spherical  triangle  are  60°,  80°,  120°,  find  the 
angles  of  its  polar  triangle. 

2.  If  the  angles  of  a  spherical  triangle  are  72°,  104°,  88°,  find  the 
sides  of  the  polar  triangle. 

3.  If  a  triangle  is  isosceles,  prove  that  its  polar  triangle  is  isosceles. 

4.  If  each  side  of  a  spherical  triangle  is  a  quadrant,  describe  its 
polar  triangle. 

5.  In  case  each  side  of  spherical  triangle  ABC  is  a  quadrant,  show 
that  the  eight  triangles  formed  by  drawing  the  great  circles  whose 
poles  are  A,  B,  and  C  are  all  congruent. 

6.  Show  that  for  any  triangle  the  construction  of  the  polar  triangle 
gives  four  pairs  of  symmetrical  triangles. 

7.  If  a  triangle  ABC  is  isosceles,  show  that  of  the  eight  triangles 
of  Ex.  5  there  are  four  pairs  of  congruent  triangles. 

8.  If  the  triangle  ABC  is  not  isosceles,  show  that  of  the  pairs  of 
triangles  proved  congruent  in  Ex.  7  none  are  now  congruent. 

9.  If  the  angles  of   a  spherical  triangle  are  70°,  80°,  and   110° 
respectively,  find  the  sides  of  each  of  the  eight  triangles  formed  by  the 
polar  construction. 

10.  Is  it  possible  to  construct  a  spherical  triangle  whose  angles 
are  50°,  60°,  120°? 

11.  Is  it  possible  to  construct  a  spherical  triangle  whose  angles 
are  60°,  120°,  150°  ? 

SUGGESTION.     Consider  the  polar  triangle  of  such  triangle. 

12.  Consider  the  questions  on  trihedral  angles  corresponding  to  the 
two  preceding. 

13.  If  the  sides  of  a  spherical  triangle  are  75°,  95°,  and  115°  re- 
spectively, find  the  angles   of  each   triangle  formed  by   the  polar 
construction. 

14.  How  can  the  theorem  of  §  257  be  used  to  prove  that  a  side  of 
a  spherical  triangle  cannot  be  as  great  as  a  semicircle  ? 

15.  If  it  is  given  that  a  spherical  triangle  is  equilateral,  can  we 
infer  from  the  theorems  thus  far  proved  that  its  polar  triangle  is 
equilateral  ? 


140 


SOLID   GEOMETRY. 


POLYHEDRAL  ANGLES    AND    SPHERICAL  POLYGONS. 

271,  THEOREM.     The  sum  of  the  face  angles  of  any 
convex  polyhedral  angle  is  less  than  four  right  angles. 


n 


IB 


Proof :  Let  ABCDE  be  a  polygonal  section  of  the  given 
polyhedral  angle.  The  number  of  triangles  thus  formed 
having  P  for  a  vertex  is  equal  to  the  number  of  face  angles 
of  the  polyhedral  angle. 

Let  O  be  any  point  in  the  base,  and  draw  (X4,  OB,  O<7,  etc. 

Then  Z  PBA  +  Z  PBC  >  Z  ABC,  and  Z  PCB  +  Z  PCD  > 
Z  £CD,  and  so  on.  (Why?) 

But  the  sum  of  the  A  of  the  A  OAB,  OBC,  etc.,  is  equal 
to  the  sum  of  the  A  of  the  A  PAB,  PBC,  etc. 

Hence,  Z  APB  +  Z  .BPC+  •••  <  Z  ^os  +  Z^ocH — . 

But  the  sum  of  the  A  about  O  is  four  right  angles. 

Therefore,  the  sum  of  the  face  angles  of  the  polyhedral 
angle  is  less  than  four  right  angles. 

272,  Definition.  The  section  of  a  sphere  made  by  a 
convex  polyhedral  angle  whose  vertex  is  the  center  of 
the  sphere  is  called  a  spherical  polygon. 

Since  a  plane  may  be  passed  through 
the  vertex  of  a  polyhedral  angle  such 
that  the  polyhedral  angle  lies  entirely 
on  one  side  of  it,  it  follows  that  a 
spherical  polygon  lies  within  one  hemi- 
sphere. 


THE  SPHERE.  141 

273,  COROLLARY.     State  and  prove  the  theorem  on 
spherical  polygons  which  corresponds  to  that  of  §  271. 

274,  THEOREM.     The  sum  of  the  angles  of  a  spherical 
polygon  of  n  sides  is  greater  than  2  (n  —  2)  right  angles 
and  less  than  2  n  right  angles. 

Proof  :    Divide  the  polygon  into  n  —  2  triangles. 

Hence,  by  §  257  the  sum  of  the  angles  is  greater  than 
2(w  —  2)  right  angles. 

Since  the  polygon  has  n  angles,  and  since  each  angle  is 
less  than  two  right  angles,  it  follows  that  their  sum  is  less 
than  2  n  right  angles. 

275,  COROLLARY.     State  and  prove  the  theorem  on 
polyhedral  angles  which  corresponds  to  the  preceding. 

AREAS  OF  SPHERICAL  POLYGONS. 

276,  Definitions.     A  spherical  polygon  divides  a  sphere 
into  two  parts,  an  exterior  and  an  interior,  so  that  every 
path  on  the  sphere  passing  from  one  to  the  other  must 
cross  the  polygon. 

We  have  seen  (§  272)  that  any  spherical  polygon  lies 
within  one  hemisphere.  The  interior  is  that  one  of  the  two 
parts  which  lies  entirely  within  this  hemisphere. 

Two  spherical  polygons  are  said  to  inclose  equal  areas 
or  to  be  equal  if  they  are  congruent,  or  if  they  can  be 
divided  into  polygonal  surfaces  which  are  congruent  in 
pairs. 

The  terms  spherical  triangle,  spherical  polygon,  are  sometimes 
used  to  refer  to  the  part  of  the  sphere  inclosed  by  these  figures.  The 
context  will  always  indicate  clearly  in  which  sense  they  are  used. 


142 


SOLID  GEOMETRY. 


277,  THEOREM.     Two    symmetrical    spherical    tri- 
angles are  equal. 


Proof :  Let  ABC  and  A'B'C'  be  the  given  triangles. 

Extend  the  radii  OA,  OB,  OC  to  meet  the  sphere  in  Av 
Bv  Cj,  thus  forming  a  triangle  symmetrical  to  A  ABC 
(§  233),  and  hence  congruent  to  A  A'B'C'  (§  243). 

Let  P  be  a  pole  of  the  circle  through  A,  B,  C.     Extend 

PO  to  meet  the  sphere  in  Pr     Draw  PA,  PB,  PC,  and  P^, 

P!-/?!,  and  PjCj. 

Suppose  that  P  lies  within  A  ABC. 

Now  prove  that  A  PAB  ^  A  P1A1BV  A  PAC  ^  A  P^Cj, 
A  PBC  ^  Pj-BjCj.  Note  that  these  triangles  are  isosceles. 

Hence,  show  that  A  ABC  =  A  ^11B1C1,  and  therefore 
AABC=AAfBfCr. 

278.  Definitions.  A  lune  is  a  figure  formed  by  two 
great  semicircles  having  the  same  end-points.  The  angle 
between  these  semicircles  is  the  angle  of  the  lune. 

A  birectangular  spherical  triangle  is  one  having  two 
right  angles. 

If  one  angle  of  a  birectangular  triangle  is  1°,  the  triangle 
incloses  one  of  720  equal  parts  of  the  sphere.  The  area 


THE  SPHERE.  143 

inclosed  by  such  a  triangle  is  called  a  spherical  degree  and 
is  used  as  a  unit  of  measure  of  areas  inclosed  by  spherical 
polygons. 

In  a  similar  manner  we  define  a  spherical  minute  and  a 
spherical  second. 

279,  THEOREM.     The  area  inclosed  ~by  a  lime  in  terms 
of  spherical  degrees  is  twice  the  angle  of  the  lune. 

• 

280,  Definition.     The  number  of  spherical  degrees  by 
which  the  sum  of  the  angles  of  a  spherical  triangle  exceeds 
180°  is  called  the  spherical  excess  of  the  triangle. 

281,  THEOREM.     The  area  of  a  spherical  triangle  in 
terms  of  spherical   degrees  is   equal   to  its  spherical 
excess. 

Proof  :  We  are  to  show  that  area 
A  ABC  =  ^A  +  Z.B+Z.C-  180°. 

Consider  the  lunes  ACDB,  CAEB, 
BCFA. 

We  have 

A  ABC  +  A  BCD  =  ACDB  —  ^/.A. 

A  ABC  +  A  BAE  =  CAEB  =  2  Z  C. 

A  ABC  +  A  CFA  =  BCFA  =  2  Z  B. 

Hence,  adding,  3  A  ABC  +  A  BCD, 
BAE,  CFA  =  2(Z  A  +  ^B  +  ^C). 

Now  A  BCD  and  AEF  are  symmetrical  and  have  equal 
areas. 

Hence, 

2A^J5C-f  A  ABC,  AEF,  BAF,  CFA  =  2(Z  A  +  Z.  B  + /.  C). 

But  A  ABC,  AEF,  BAF,  CFA  together  constitute  a  hemi- 
sphere or  360  spherical  degrees. 

Hence,  2  A  ABC  +  360°  =  2(Z  A  4-  ^  B  +  Z.  C). 

Solving,  A  ABC=  /.  A  +  Z.  B  +  Z  C-  180°. 


144  SOLID  GEOMETRY.. 

282,  Definition.      The    spherical   excess   of   a   spherical 
polygon  is  the  sum  of  its  angles  less  (n  —  2)180°,  where  n 
is  the  number  of  sides  of  the  polygon. 

283,  THEOREM.     The   area  of  a  spherical  polygon 
in  terms  of  spherical  degrees  is  equal  to  its  spherical 

excess. 

Proof :  Join  one  vertex  of  thfc  polygon  to  each  non- 
adjacent  vertex,  thus  forming  n  —  2  spherical  triangles. 

Now  prove  that  the  sum  of  the  spherical  excesses  of 
these  triangles  is  the  spherical  excess  of  the  polygon  and 
thus  complete  the  proof. 

284,  EXERCISES. 

1.  What  is  the  area  in  spherical  degrees  of  a  birectangular  tri- 
angle one  of  whose  angles  is  54°?     If  one  angle  is  79°  30' ;  106°;  14' ; 

36"  ? 

2.  What  is  the  area  inclosed  by  a  lune  whose  angle  is  45°?     Note 
that  the  lune  may  be  divided  into  two  birectangular  triangles.     What 
is  the  third  angle  of  each  ? 

3.  Between  what  limits  is  the  sum  of  the  angles  of  a  spherical 
polygon  of  eight  sides  ? 

4.  If  the  sum  of  the  angles  of  a  spherical  polygon  is  11  right 
angles,  what  is  known  about  the  number  of  its  sides? 

5.  If  the  sum  of  the  angles  of  a  spherical  polygon  is  14  right 
angles,  what  is  known  about  the  number  of  its  sides  ? 

6.  The  sides  of  a  spherical  ^nljii^m  are  85°,  95°,  110°.     Find  the 
area  of  each  of  the  eight  triangles  formed  by 'the  polar  construction 
from  this  triangle. 

7.  The  area  of  a  spherical  triangle  is  74  spherical  degrees.     One 
angle  is  105°.     Of  the  other  two  angles  one  is  twice  the  other.     Find 
all  the  angles  of  the  triangle. 


.  3 

THE  SPHERE.  145 

AREA  AND  VOLUME  OF  THE  SPHERE. 

285,  In  §§  276-283  we  discussed  the  areas  inclosed  by 
spherical  polygons  in  terms  of  a  unit  directly  applicable  to 
the  sphere,  namely,  the  spherical  degree. 

We  now  consider  the  area  of  the  sphere  in  terms  of  a 
plane  unit  of  measure.  It  is  clear  that  such  measurement 
can  be  approximate  only,  since  no  plane  segment  however 
small  will  coincide  with  the  spherical  surface.  Similarly, 
if  the  cube  is  used  as  the  unit  of  volume,  the  measurement 
of  the  volume  inclosed  by  a  sphere  must  be  approximate, 
since  no  set  of  cubes  however  small  can  be  made  exactly  to 
coincide  with  a  sphere. 

The  two  following  theorems  are  needed : 

286,  THEOREM.     The  lateral  area  of  a 
frustum  of  a  right  circular  cone  is  equal 
to  the  altitude  of  the  frustum  multiplied 
by  the  length  of  a  circle  whose  radius  is 

the  perpendicular  distance  from  a  point  A        F 

in  the  axis  of  the  frustum  to  the  middle  point  of  an  . 

element. 

Given  A  A'  an  element  of  the  frustum  whose  middle  point  is  Bl, 
CC'  the  axis  of  the  frustum,  BD  J_  CC'  and  EB±  A  A'. 

To  prove  that  the  lateral  area  is  equal  to  2  TT  •  EB  •  CCr. 
Proof:  By  §  164,  the  lateral  area  is  2-Tr  •  BD  -  AAf. 
Hence,  we  must  show -that  EB  •  CCf  =  BD  •  AAf. 
To  do  this,  draw  A' FA.  ^LCand  show  that  A  AFA'  ~AtEDB. 

287,  COROLLARY.     The  lateral  area  of  a  cone' is  equal 
to  its  altitude  times  the  length  of  a  circle  whose  radius 
is   the  perpendicular  from  a  point  in  the  axis  to  the 
middle  point  of  an  element. 


146 


SOLID  GEOMETRY. 


288,  THEOREM.  Given  a  fixed  line  through  the  ver- 
tex of  a  triangle,  but  not  crossing  it.  (  The  volume 
swept  out  ~by  the  triangle  as  it  rotates  about  the  fixed 
line  as  an  axis  is  equal  to  the  numerical  measure  of  the 
area  generated  %  the  side  opposite  the  fixed  vertex 
multiplied  by  one  third  the  altitude  upon  that  side. 


FIG.  1 


FIG.  2 


Given"  A  ABC  with  vertex  A  in  the  fixed  line  /. 


CASE  1.      When  one  side  as  AB  lies  in  line  I.     (Fig.  1.) 
Draw  AD  _L  BC,  or  BC  produced,  and  CE  A.  AB. 
Then    Vol.  ABC  =  Vol.  AEG  +  Vol.  BEG 

=  ^C3*-AB.      (Why?) 
6 

But  OE?.AB=CE'  CE-AB  =  CE'BC-AD.     (Why?) 

And  by  §  158,  IT  •  CE  •  BC  is  the  area  swept  out  by  BC. 
Hence,  Vol.  ABC  =  ^  AD  -  (area  generated  by  JBC). 

CASE  2.     When  neither  AB  nor  AC  lies  in  line  I.    (Fig.  2.) 
Produce  CB  to  meet  Im  F  and  draw  AD  and  CE  as  before. 


THE  SPHERE. 


147 


Then  Vol.  ABC  =  Vol.  AFC  —  Vol.  AFB. 

By  Case  1  Vol.  AFC—  \AD  -  (area  generated  by  FC), 
and  Vol.  AFB  =  ^  AD  •  (area  generated  by  FB). 

Hence,  Vol.  ABC  =  J  AD  •  (area  generated  by  FC  —  FB*), 
or  Vol.  ABC  =  J  4D  •  (area  generated  by  JBC). 

Let  the  student  give  the  proof  when  the  triangle  is  of  other  forms, 
e.g.  in  Case  1  when  AD  falls  ivithin  the  triangle,  and  in  Case  2  when 
EC  is  parallel  to  I. 

289,  Area  of  the  sphere.  About  a  circle  circumscribe  a 
polygon  as  follows :  Construct  two  diameters  AB  and  CD 
at  right  angles  to  each  other  and  divide  each  quadrant 
into  an  even  number  of  parts  by  points,  as  D,  E,  F.  At 
each  alternate  division  point, 
beginning  with  the  first  point 
Z>,  draw  a  tangent. 

There  results  a  regular  poly- 
gon with  vertices  on  the  diam- 
eters AB  and  CD. 

If  now  we  construct  another 
polygon  in  the  same  manner 
by  dividing  each  quadrant  into 
twice  as  many  arcs,  it  will  like- 
wise have  two  vertices  on  each 
of  the  diameters  AB  and  <7D,  and  this  may  be  repeated  at 
pleasure. 

Now  inscribe  a  polygon  similar  to  the  one  just  circum- 
scribed by  joining  the  points  C  and  E,  E  and  A,  and  so  on. 

If  now  the  whole  figure  is  made  to  revolve  about 
AB  as  an  axis,  the  circle  generates  a  sphere,  and  the 
circumscribed  and  inscribed  polygons  generate  sets  of 
circumscribed  and  inscribed  cones  and  frustums  of  cones. 

We  assume  the  following : 


148 


SOLID   GEOMETRY. 


290,  Axiom  IX.  A  sphere  has  a  definite  area  and 
incloses  a  definite  volume  which  are  less  respectively 
than  the  volume  and  the  surface  of  any  circumscribed 
figure  and  greater  than  those  of  any  inscribed  convex 
figure. 


291,    THEOREM.     Tlie  area  of  a  sphere  is  4 

Proof  :  Denote  by  r  the  radius  of  the  circle  which  gener- 
ates the  sphere.     That  is,  in  the  figure,  r  =  OF=  OD  =  OG. 


H 


Then  by  §  286  the  lateral  areas  of  the  frustums  whose 
axes  are  OL  and  OK  are  2  TTT  -  OL  and  2  Trr  -  OK,  and  by 
§  287  the  lateral  areas  of  the  cones  whose  axes  are  LN  and 
KM  are  respectively  2  Trr  •  LN  and  2  Trr  •  KM. 

Hence,  the  total  surface  of  the  whole  circumscribed  figure 
is 


or 


2  Trr  •  MK  -h  2  Trr  •  KO  +  2  TTT  •  OL  +  2  Trr  -  LN, 
2  TTT  (MK  +  KO  +  OL  +  LN)  =  2  TTT  -  MN. 


If  now  a  polygon  of  twice  the  number  of  sides  is  con- 
structed, in  a  manner  similar  to  the  above,  we  obtain 
another  circumscribed  figure  whose  area  is  2  Trr  •  M '  N1 
where  M '  and  Nf  are  the  two  vertices  on  the  line  AB. 


THE  SPHERE.  149 

As  this   process  is   repeated  indefinitely,  the    vertices 
which  lie  on  the  line  AB  may  be  made  to  approach  as  near 
as  we  please  to  A  and  .B,  and  hence  the  total  surface  gen- 
erated approaches  as  near  as  we  please  to 
A  =  2  Trr  -  AB  =  ±Trr2. 

We  now  prove  that  A  cannot  differ  from  4  Trr2. 

(a)  Suppose  A  >  4  Trr2,  and  let  A  =  4  Trr2  +  d.  (1) 

Let  k  be  the  difference  between  M'N'  and  AB. 

Then  the  area  of  the  circumscribed  figure  is 
2  TTT  (2  r  +  k)  =  4  Trr2  +  2  Trr&. 

Since  &  can  be  made  as  small  as  we  please,  2  Trrk  can 
be  made  smaller  than  d. 

Hence,  by  Axiom  IX,  the  area  >  4  Trr2  +  d,  which  con- 
tradicts the  hypothesis  (1). 

Therefore  the  area  cannot  be  greater  than  4  Trr2. 

(5)  In  the  inscribed  figure  let  OF'  be  the  apothem. 

Then  by  §  286  and  an  argument  like  that  used  above, 
we  find  that  the  area  of  the  inscribed  figure  is 

2  TT  •  AB  •  OF'  =  4  Trr  •  OF'. 

By  continuing  to  double  the  number  of  sides,  OF'  may 
be  made  to  approach  as  nearly  as  we  please  to  OF  =  r. 

Suppose  that  A  <  4  Trr2  and  let  A  =  4  Trr2  -  d'.  (2) 

Let  Jc1  be  the  difference  between  r  and  OF'. 

Then  the  area  of  the  inscribed  figure  is 

4  Trr  -  OF'  =  4  Trr(r  -  k')  =  4  Trr2  -  4  Trrk'. 

Since  k'  can  be  made  as  small  as  we  please,  4-7rr&'  can 
be  made  less  than  d' .  Hence  there  is  some  inscribed  fig- 
ure whose  area  is  greater  than  4  Trr2  —  d' .  Hence  by 
Axiom  IX,  A  >  4 Trr2-  d',  which  contradicts  (2). 

Therefore  the  area  cannot  be  less  than  4  Trr2.  Since  the 
area  is  neither  less  nor  greater  than  4  Trr2,  it  is  therefore 
equal  to  4  Trr2. 


150 


SOLID   GEOMETRY. 


292.   THEOKEM.     The  volume  of  a  sphere  is  f  ^. 

Proof :  From  §  288  the  volume  of  the  figures  formed  by 
rotating  the  triangles  O1H  and  OHN  about  AB  as  an  axis 


is 


-  •  (area  generated  by  IH 
o 


H 


D 


That  is,  the  volume  of  the  whole  circumscribed  figure 

is  equal  to  the  surface  of  the  figure  multiplied  by  — . 

3 

Now  suppose  the  volume  of  the  sphere  not  equal  to 
|  Trr3,  but  equal  to  f  Trr3  +  d. 

Since  by  the  argument  of  §  291  we  can  obtain  a  circum- 
scribed figure  whose  surface  is  as  near  as  we  please  to 
4  ?rr2,  it  follows  that  one  can  be  found  whose  volume  is 
less  than  ^  Trr3  +  d. 

Hence,  the  volume  cannot  be  greater  than  |  Trr3. 

In  a  similar  manner,  using  the  inscribed  figures  show 
that  the  volume  cannot  be  less  than  -|  Trr3. 

293,  The  volume  of  a  sphere  by  inspection.  The  fact 
that  the  volume  of  a  sphere  is  equal  to  its  area  multiplied 
by  one  third  its  radius  may  be  inferred  directly  from  the 
accompanying  figure- 


THE  SPHERE.  151 

The  sphere  is  covered  with  a  network  of  small  spher- 
ical quadrilaterals.     If  these  are  taken 
small  enough,  they  may  be  regarded  as 
approximately  plane  surfaces. 

On  this  supposition  we  have  a  set  of 
pyramids  with  a  common  altitude  r  and 
the  sum  of  their  bases  approximately 
equal  to  the  area  of  the  sphere. 

Hence,  their  combined  volume  is  -J  r  x  (area  of  sphere) 
or  i  r  •  4  77T2.  That  is,  the  volume  is  ^  Trr3. 

It  is  clear  that,  by  making  these  quadrilaterals  suffi- 
ciently small,  this  result  may  be  approximated  as  nearly  as 
we  please. 


294,  EXERCISES. 

1.  The  surface  of  a  polyhedron  circumscribed  about  a  sphere  of 
radius  4  inches  is  420  square  inches.     Find  its  volume. 

2.  The  volume  of  a  polyhedron  circumscribed  about  a  sphere  of 
radius  3.5  inches  is  450  cubic  inches.     Find  its  surface. 

3.  Given  a  sphere  of  radius  6  inchests  there  any  upper  limit  to 
the  volume  of  its  circumscribed  polyhedrons  ?    That  is,  can  polyhe- 
drons be  circumscribed  having  a  volume  as  large  as  we  please? 

4.  On  the  same  sphere  is  there  any  lower  limit  to  the  volume  of 
its  circumscribed  polyhedrons  ? 

5.  Show  that  the  areas  of  two  spheres  are  in  the  same  ratio  as 
the  squares  of  their  radii  or  of  their  diameters. 

6.  Show  that  the  volumes  of  two  spheres  are  in  the  same  ratio  as 
the  cubes  of  their  radii  or  of  their  diameters. 

7.  Find  the  area  of  a  sphere  whose  radius  is  8  inches. 

8.  Find  the  volume  of  a  sphere  whose  radius  is  10  feet. 

9.  If  the  area  of  a  sphere  is  227  sq.  ft.,  find  its  radius. 

10.  If  the  volume  of  a  sphere  is  335  cu.  in.,  find  its  radius. 

11.  If  the  volumes  of  two  spheres  are  27  cu.  in.  and  729  cu.  in., 
compare  their  radii. 


152 


SOLID   GEOMETRY. 


295,  Definition.  That  part  of  a  sphere  included  between 
two  parallel  planes  cutting  it  is  called  a  zone.  The  per- 
pendicular distance  between  the  planes  is  the  altitude  of 
the  zone. 

The  figure  formed  by  a  zone,  together  with  the  circular 
plane-segments  cut  out  by  the  sphere,  is  called  a  spheri- 
cal segment,  and  the  circular  plane-segments  are  its  bases. 

If  one  of  the  cutting  planes  is  tangent  to  the  sphere, 
then  the  spherical  segment  and  the  corresponding  zone 
are  said  to  have  but  one  base.  The  altitude  is  the  perpen- 
dicular distance  from  the  base  to  the  tangent  plane. 

If  one  nappe  of  a  convex  conical  surface 
has  its  vertex  at  the  center  of  a  sphere,  the 
portion  cut  off  by  the  sphere,  together  with 
the  intercepted  part  of  the  spherical  surface, 
is  called  a  spherical  cone. 

If  two  spherical  cones  have  the  same 
axis,  one  lying  within  the  other,  the  figure 
formed  by  their  two  lateral  surfaces,  to- 
gether with  the  part  of  the  sphere  inter- 
cepted between  them,  is  called  a  spherical 
sector. 

If  the  two  cones  are  right  circular  cones, 
they  intercept  circles  on  the  sphere,  and  the 
zone  thus  included  is  called  the  base  of  the 
spherical  sector. 

If  the  accompanying  figure  be  revolved  about  LAfas 
an  axis,  then  any  arc,  as  GD  or  MD,  generates  a  zone, 
the  former  with  two  bases,  the  latter  with  one. 

The  figure  MDF  or  FDGE  generates  a  spherical  segment,  the 
former  with  one  base,  the  latter  with  two. 

The  figure  CA L  or  CBL  generates  a  spherical  cone,  CL  being  the 
common  axis. 

The  figure  CBA  generates  a  spherical  sector. 


M 


THE  SPHERE.  153 

296,  Area  of  a  zone.     An  argument  precisely  like  that 
of  §  291  shows  that  the  area  of  a  zone  is 

A  =  2  *rh 

where  h  is  the  altitude  of  the  zone. 

That  is,  instead  of  AB,  the  diameter  in  case  of  the 
sphere,  we  should  have  the  sum  of  the  altitudes  of  the 
frustums  circumscribed  about  the  zone  equal  to  A,  the  alti- 
tude of  the  zone. 

297.  Volume  of   a  spherical  cone.     An    argument   pre- 
cisely like  that  of  §  292  shows   that   the   volume    of   a 
spherical  cone  is  v—r 

-3' 

where  A  is  the  area  of  the  zone  cut  out  of  -the  sphere  by 
the  cone.     Hence,  if  h  is  the  altitude  of  this  zone,  we  have 


. 
3  3 

In  like  manner  the  volume  of  a  spherical  sector  is 


298,  EXERCISES. 

1.  The  radius  of  a  sphere  is  6  and  the  altitude  of  a  zone  is  5. 
Find  the  area  of  the  sphere  and  of  the  zone. 

2.  The  area  of  a  zone  is  36  TT  and  its  altitude  4.     Find  the  radius 
of  the  sphere. 

3.  If  the  radius  of  a  sphere  is  r,  find  the  perpendicular  distance 
from  a  point  A  to  the  plane  of  a  given  great 

circle  if  the  distance  on  the  sphere  from  A  to 
the  great  circle  is  35°. 

4.  Solve  the  preceding  problem  if  the  dis- 
tance on  the  sphere  from  A  to  the  given  great 
circle   is  23£°.     Also   if   the  distance  is  66^°. 
(Use  the  tables,  page  57.) 


154 


SOLID   GEOMETRY. 


299,   PROBLEM.     To  find  the  volume  of  a  spherical 
segment. 

Solution.  Let  r  be  the  radius  of 
the  sphere,  and  r-^  and  r2  the  radii  of 
the  bases  of  the  segment,  a  the  altitude 
of  the  zone,  and  let  the  segment  be 
generated  by  revolving  the  figure  ACDB 
about  AO  as  an  axis. 


We  have  Vol.  generated  by  ODB  =  —  r2a. 

o 

Vol.  generated  by  OAB  =  ^-r^ 

Vol.  generated  by  OCD  =  ~r^ 

o 

Hence,      F  =  ^r2a  +  ^r^( 


From 
we  obtain 


r-r-a* 


§  297 
(Why?) 
(Why?) 

(1) 
(2) 


Substituting  this  value  of  d  in  r2  =  r22  +  c?2, 

_  r  4  +  r  4  +  ^4_  2  r  2r  2  +  2  q2r22  +  2  ^ 


we  get 


4o2 


(3) 


Substituting  (2)  and  (3)  in  (1)  and  reducing,  we  have 


Hence,  we  have  the  result  : 

THEOREM.  The  volume  of  a  spherical  segment  is 
numerically  equal  to  the  sum  of  the  areas  of  its  bases 
multiplied  by  half  its  altitude,  plus  the  volume  of  a 
sphere  whose  diameter  is  equal  to  the  altitude  of  the 
segment. 


THE  SPHERE.  155 

SUMMARY  OF  CHAPTER  V. 

1.  Collect  the  theorems  involving  plane  sections  of  the  sphere. 

2.  Collect  the  definitions  involving  plane  sections  of  the  sphere. 

3.  State  some  of  the  principal  exercises  and  problems  connected 
with  plane  sections  of  the  sphere. 

4.  Collect  the  definitions  on  trihedral  angles  and  spherical  tri- 
angles. 

5.  Arrange  in  parallel  columns  the  pairs  of  theorems  on  trihedral 
angles  and  spherical  triangles,  which  are  proved  without  the  use  of 
polar  triangles. 

6.  Collect  the  definitions  on  polar  triangles. 

7.  Collect  the  theorems  on  polar  triangles. 

8.  Continue  the  lists  started  in  Ex.  5,  adding  the  theorems  proved 
by  means  of  polar  triangles. 

9.  Make  a  list  of  the  definitions  involving  polyhedral  angles  and 
spherical  polygons. 

10.  Collect  the  theorems  involving  polyhedral  angles  and  spherical 
polygons. 

11.  Collect  the  theorems  on  the  areas  of  spherical  triangles  and 
polygons. 

12.  Give  the  definitions  and  axioms  pertaining  to  the  area  and 
volume  of  the  sphere. 

13.  State  all  the  theorems  pertaining  to  the  area  and  volume  of 
the  sphere. 

14.  Give  the   definitions   and  theorems  pertaining  to  spherical 
figures,  such  as  zones,  cones,  sectors,  segments. 

15.  Collect  all  the  mensuration  formulas  in  this  chapter. 

16.  Collect  all  the  mensuration  formulas  of  Solid  Geometry. 

17.  Which   of  these   formulas  are  illustrations    of  the  general 
theorem  that  the  surfaces  of  similar  solids  are  in  the  same  ratio 
as  the   squares  of   their  corresponding  linear  dimensions?     Which 
ones  are  special  instances  of  the  theorem  that  the  volumes  of  similar 
solids  are.  in  the  same  ratio  as  the  cubes  of  corresponding  linear 
dimensions? 

18.  Describe  some  of   the    most   important   applications   in   this 
chapter.     Return  to  this  question  after  studying  the  following  list. 


156  rW  SOLID  GEOMETRY. 


PROBLEMS   AND   APPLICATIONS. 


1.  What  part  of  the  earth's  surface  liesr  in  the  torrid  zone  ?    What 
part  in  the!  temperate  zones  ?     What  pap  in  the  frigid  zones?     The 
parallels  2f  |°  north  and  south  of  the  equator  are  the  boundaries  of 
the  torrid   zone,   and  the  parallels  6v|°  north   and  south   are  the 
boundaries  of  the  frigid  zones. 

2.  Find  to  four  places  of  decimals  the  area  of  a  sphere  circum- 
scribed about  a  cube  whose   edge   is   6.     No  square  root  is  to  be 
approximated  in  the  process,  and  the  value  of  TT  is  taken  3.1416. 

3.  Can   the  volume  of   the  sphere  in  the  preceding  exercise  be 
approximated  without  finding  a  square  root?     Find  the  volume. 

4.  Find  the   area  of  a  sphere  circumscribed  about  a  rectangular 
parallelepiped  whose  sides  are  a,  6,  and  c. 

5.  Find  the  volume  of  the  sphere  in  the  preceding  example. 

6.  A  fixed  sphere  with  center  0  has  its  center  on  another  sphere 
with  center  0'.     Show  that  the  area  of  the  part  of  0'  which  lies 
within  0  is  equal  to  the  area  of  a  great  circle 

of  the  sphere  0,  provided  the  radius  of  the 
sphere  0  is  not  greater  than  the  diameter  of  0'. 
SUGGESTION.  Let  the  figure  represent  a 
cross  section  through  the  centers  of  the  two 
spheres.  Connect  O  with  A  and  R.  Then 
O42  =  OB  x  OD.  But  OD  is  the  altitude  of  the  zone  of  O1,  which 
lies  within  0,  and  OB  is  the  diameter  of  the  sphere  0'.  Hence,  the 
area  of  the  zone  is  TT  BO  x  OD  =  TT  OA2. 

7.  Given  a  solid  sphere  of  radius  12  inches.     A  cylindrical  hole 
is  bored  through  it  so  that  the  axis  of  the  cylinder  passes  through  the 
center  of  the  sphere.     What  area  of  the  sphere  is  removed  if  the 
diameter  of  the  hole  is  4  inches? 

8.  Find   the   volume   removed   from  the   sphere   by  the   process 
described  in  the  preceding  exercise. 

9.  Through  a  sphere  of  radius  r  a  hole  of  radius  rl  is  bored  so 
that  the  axis  of   the  cylinder  cut  out  passes  through  the  center  of 
the  sphere.     Find  the  volume  of  the  sphere  which  remains. 


THE  SPHERE.  157 

10.  A  cylindrical  post  6  in.  in  diameter  is  surmounted  by  a  part 
of  a  sphere  10  in.  in  diameter,  as  shown  in  the  figure. 

Find  the  surface  and  the  volume  of  the  part  of  the 
sphere  used. 

11.  A   cylindrical  post  5  ft.  long   and   4   in.   in 
diameter  is  turned  so  as  to  leave  on  it  a  part  of  a 
sphere  7  in.  in  diameter  and  having  its  center  in  the 
axis  of  the  post.     Find  the  volume  of  the  whole  post. 

12.  Find  the  volume  of  a  spherical  shell  1  inch 
thick  if  its  outer  diameter  is  8  inches. 

13.  What  is  the^iameter  of  a  spherical  shell  an  inch  thick  whose 
volume  is  half  that  inclosed  within  «usphere 

14.  Compare  the  volumes  and  areas  of  a  sphere  and  the  circum- 
scribed cylinder. 

15.  In  a  sphere  of  radius  r  inscribe  a  cylinder  whose  altitude  is 
equal  to  its  diameter.     Compare  its  volume  and  area  with  those  of 
the  sphere. 

16.  In  a  sphere  inscribe  a  cylinder  whose  altitude  is  n  times  its 
diameter.     Compare  its  area  and  volume  with  those  of  the  sphere. 

17.  Compute  the  length  of  the  diagonal  of  a-  cube  in  terms  of  its 
side,  and  also  the  length  of  the  side  in  terms  of  half  the  diagonal. 

18.  Express  the  volume  of  a  cube  inscribed  in  a  sphere  in  terms 
of  the  radius  of  the  sphere. 

19.  Three  spheres  each  of  radius  r  are- placed  on  a  plane  so  that 
each  is  tangent  to  the  other  two.     A  fourth  sphere  of  radius  r  is 
placed  on  top  of  them.     Find  the  distance  from  the  plane  to  the  top 
of  the  upper  sphere. 

20.  Find  the  vertical  distance  from  the 
floor  to  the  top  of  a  triangular  pile  of  spher- 
ical cannon  balls,  each  of  radius  5  inches, 
if  there  are  3  layers  in  the  pile. 

21.  Solve  a  problem  like  the  preceding 
if  there  are  16  layers  in  the  pile,  each  shot 
of  radius  r. 

22.  Solve  a  problem  like  the  preceding 
for  a  square  pile  of  shot  of  12  layers. 

23.  Solve  Exercises  21  and  22  if  there  are  n  layers  in  each  pile. 


CHAPTER   VI. 
VARIABLE  GEOMETRIC  MAGNITUDES, 

GRAPHIC  REPRESENTATION. 

300,  It  is  often  useful  to  think  of  a  geometric  figure  as 
continuously  varying  in  size,  or  in  shape,  or  both. 

E.g.  if  a  parallelopiped  has  a  fixed  base,  say  24  square  inches,  but 
an  altitude  which  varies  continuously  from  3  inches  to  5  inches,  then 
the  volume  varies  continuously  from  3  •  24  =  72  to  5  •  24  =  120  cubic 
inches. 

We  may  even  think  of  the  altitude  as  starting  at  zero  inches  and 
increasing  continuously,  in  which  case  the  volume  also  starts  at  zero. 

From  this  point  of  view  many  theorems  may  be  repre- 
sented graphically.  The  graph  has  the  advantage  of  ex- 
hibiting the  theorem  at  once  for  all  its  particular  cases. 

For  a  description  of  graphic  representation,  see  Chapter  V  of  the 
authors'  High  School  Algebra,  Elementary  Course. 

301,  THEOREM.     The  volumes  of  two  parallelopipeds 
having   equal   bases  are  in    the   same   ratio   as    their 
altitudes.* 

Graphic  Representation.     By  §  86  we  have 

Volume  =  base  x  altitude. 

Consider  parallelopipeds  each  with  a  base  whose  area  is  A,  and  with 
altitudes  hi,  h2,  ha,  and  corresponding  volumes  Fi,  F2,  T73. 

Then          £=7^=4-  £=4r!  =  r2>ete-  (1) 

Vz      Ah2      h2      Vz      Ah%      hz 

Consider  the  case  where  A  =  10.  Let  one  horizontal  space  repre- 
sent one  unit  of  altitude,  and  one  vertical  space  ten  units  of  volume. 

Thus,  the  point  P2  has  the  ordinate  V  —  10  vertical  units  (repre- 
senting 100  units  of  volume),  and  the  abscissa  h%=  10  horizontal  units. 

158 


VARIABLE  GEOMETRIC  MAGNITUDES. 


159 


Similarly,  locate  the  points  PI  and  PS. 

Using  equations  (1),  show  that  0,  PI,  P2,  PS  lie  in  a  straight  line. 


mo 


D  Altitude —Axis 


If  we  suppose  that  while  the  base  of  the  parallelopiped 
remains  fixed,  the  altitude  varies  continuously  through  all 
values  from  h2  =  10  to  hB  =  20  =  2  x  10,  then  the  volume 
varies  continuously  from  10  x  10  =  100  to  10  x  20  =  200. 

Using  any  altitude  as  an  abscissa  and  the  corresponding 
volume  as  an  ordinate,  show,  as  in  §  368,  Plane  Geometry, 
that  the  point  so  determined  lies  on  the  line  OP2P3. 

302,    The  preceding  theorem  may  also  be  stated: 

The  volume  of  a  parallelopiped  with  a  fixed  base 
varies  directly  as  its  altitude. 

This  means  that  if  V  and  h  are  the  varying  volume  and  altitude, 
and  Fj  and  h±  the  volume  and  altitude  at  any  given  instant,  then 

—  =  —  or  V  =  —  •  h,  or  V  =  kh,  where  k  is  thejixed  ratio  ^JL. 
V}      hi  h^  hi 

The  graph  representing  the  relation  of  two  variables 
when  one  varies  directly  as  the  other  is  a  straight  line. 


160 


SOLID   GEOMETRY. 


303,  PROBLEM.  To  represent  graphically  the  rela- 
tion between  the  area  and  the  edge  of  a  cube  as  the  edge 
varies  continuously. 


49 


25 


16 


Axis  for  $ide$ 


Solution.  On  the  horizontal  axis  lay  off  segments  equal 
to  the  various  values  of  the  edge  0,  and  on  the  vertical  axis 
lay  off  segments  equal  to  the  corresponding  areas  A. 


VARIABLE  GEOMETRIC  MAGNITUDES.  161 

If  one  vertical  space  represents  one  unit  of  area,  then 
the  points  PJ,  P2,  P3,  etc.,  lie  on  the  steep  curve. 

The  student  should  locate  many  more  points  between 
those  here  shown,  and  see  that  a  smooth  curve  can  be 
drawn  through  them  all. 

304,  The  graph  of  the  relation  between  two  variables, 
one  of  which  varies  as  the  square  of  the  other,  is  always 
similar  to  the  one  just  given. 

The  lowest  curve  represents  the  relation  between  the 
side  of  a  square  and  its  area,  and  the  third  curve  rep- 
resents the  relation  between  the  edge  of  a  regular  tetra- 
hedron and  its  total  surface. 

In  each  of  these  cases  the  area  is  said  to  vary  as  the 
square  of  the  side  of  the  figure.  These  are  special  cases 
of  the  theorem,  §  199,  that  the  surfaces  of  similar  figures 
are  in  the  same  ratio  as  the  squares  of  their  corresponding 
linear  dimensions.  This  theorem  may  also  be  stated : 

The  areas  of  similar  figures  vary  as  the  squares  of 
their  linear  dimensions. 


305.  EXERCISES. 

1.  From  the  graph  find  approximately  the  area  of  a  regular  tetra- 
hedron whose  side  is  2.5;  one  whose  side  is  3.7;  4.3. 

2.  Find  approximately  from  the  graph  the  edges  of  cubes  whose 
total  areas  are  25  square  units ;  40  square  units ;  56  square  units. 

3.  Find  approximately  from  the  graph  the  edges  of  regular  tetra- 
hedrons whose  total  areas  are  42  square   inches ;  17  square  inches ; 
55  square  inches. 

4.  Construct  a  graph  showing  the  relation  between  the  edge  and 
the  total  area  of  a  regular  octahedron. 


162 


SOLID   GEOMETRY. 


306,   PROBLEM.     To  construct  a  graph  showing  the 
relation  between  the  edge  of  a  cube  and  its  volume. 


Axis  for  sides 


VARIABLE  GEOMETRIC  MAGNITUDES.  163 

Solution.  Taking  ten  horizontal  spaces  to  represent  one 
unit  of  length  of  side  of  the  cube  and  one  vertical  unit  to 
represent  one  unit  of  volume,  construct  the  middle  graph 
shown  on  the  page  opposite. 

307.  The  lower  graph  represents  the  relation  between 
the  length  e  of  the  edge  of  a  regular  tetrahedron  and  its 
volume  F";  and  the  upper  graph  represents  the  relation 
between  the  radius  and  the  volume  of  a  sphere. 

In  each  of  these  cases  the  volume  is  said  to  vary  as  the 
cube  of  the  given  linear  dimension. 

These  are  special  cases  of  the  theorem,  §  199,  that  the 
volumes  of  similar  solids  are  in  the  same  ratio  as  the 
cube  of  their  ratio  of  similitude. 

This  theorem  may  now  be  stated: 

The  volumes  of  similar  solids  vary  as  the  cubes  of 
their  linear  dimensions. 

308,  EXERCISES. 

1.  From  the  graph  read  off  approximately  the  cubes  of  1.3;  2.4; 
3.7. 

2.  From  the  same  graph  read  off  approximately  the  cube  roots  of 
17;  46;  54;  86. 

3.  Find  approximately  by  means  of  the  graph  the  volume  of   a 
sphere  whose  radius  is  1.5  ;  2.3 ;  2.7. 

4.  What  is  the  radius  of  a  sphere  whose  volume  is  26 ;   52 ;  71 ; 
80? 

5.  Construct  a  graph  giving  the  relation  between  the  volume  and 
the  length  of  a  side  of  regular  octahedrons. 

6.  By  means  of  the  graph  just  constructed  find  the  volume  of  a 
regular  octahedron  whose  side  is  1.3 ;  2.7 ;  3.6. 

7.  From  the  graph  constructed  under  Ex.  5  read  off  approximately 
the  lengths  of  an  edge  of  a  regular  octahedron  whose  volume  is  18 ; 
also  of  one  whose  volume  is  46. 


164 


SOLID   GEOMETRY. 


220 
210 


" 


234 
Avcis  for  sides 


VARIABLE  GEOMETRIC  MAGNITUDES.  165 

309,  The  graph  on  the  opposite  page  exhibits  the  varia- 
tion of  the  area  and  the  volume  of  the  cube.  We  notice 
that  for  small  values  of  e  the  numerical  value  of  the  area 
is  greater  than  that  of  the  volume.  For  e  =  6  these 
values  are  equal,  and  for  e  greater  than  6,  the  numerical 
value  of  the  volume  is  greater  than  that  of  the  area. 

This  is  an  instance  of  the  general  fact  that  if  a  solid 
figure  increases  in  size  but  remains  similar  to  its  first  form, 
then  after  a  certain  point  its  volume  increases  more  rapidly 
than  its  area  or  the  area  of  any  cross  section. 

310, .  The  fact  that  one  variable  y  varies  as  the  square 
or  the  cube  of  another  variable  x  is  expressed  algebraically 
by  the  equations  y  =  kx2  and  y  =  kx5,  respectively,  where 
k  is  a  constant  number  to  be  determined  in  any  particular 
case. 

If  the  value  of  k  is  known,  the  relations  represented  by 
these  equations  may  be  shown  by  a  graph  like  those  on 
the  preceding  pages. 

The  equation  y  —  kx3  may  be  plotted  by  multiplying 
the  ordinate  of  each  point  of  the  graph  V  —  e3  (see  page 
opposite)  by  the  value  of  k. 

SUMMARY  OP   CHAPTER  VI. 

1.  Give  a  brief  summary  of  the  graphic  representation  process  in 
.  general  and  in  particular,  as  applied  to  geometrical  figures. 

2.  Make  a  list  of  the  theorems  which  are  represented  graphically 
in  this  chapter. 

3.  Show  in  what  respects  such  a  representation  has  advantages  over 
the  algebraic  representation. 

4.  State  as  formulas  the  important  laws  of  variation  for  geometric 
magnitudes  discussed  in  this  chapter. 

5.  Make  a  collection  of  important  applications  given  in  this  chap- 
ter, including  those  which  occur  in  the  following  set. 


166     -••er  SOLID   GEOMETRY. 


PROBLEMS   AND   APPLICATIONS. 

1.  Assuming  that  the  weights  of  schoolboys  vary  as  the  cubes  of 
their  heights,  construct  a  graph  representing  the  relation  between  their 
heights  and  weights,  if  a  boy  5  feet  p  inches  tall  weighs  130  pounds. 

SUGGESTION.  If  w  represents  the  number  of  pounds  in  weight 
and  h  the  number  of  feet  in  height,  w  =  kh3.  From  w  —  130,  when 
h  =  5$,  we  have  k  =  .^Ar.  --  ,£>  $V 

For  the  purpose  of  the  graph,  k  =  .7  is  accurate  enough.  Hence, 
we  obtain  the  required  graph  by  multiplying  each  ordinate  of  the 
graph  of  V  =  e*  by  .7. 

2.  From  the  graph  constructed  in  the  preceding  example  find  the 
weight  of  a  boy  5  feet  tall  ;  one  5  feet  4  inches  ;  one  5  feet  6  inches. 
Compare  with  the  weights  of  boys  in  your  class. 

3.  If  a  man  6  feet  tall  weighs  185  pounds,  construct  a  graph  rep- 
resenting the  weights  of  men  of  similar  build  and  of  various  differ- 
ent heights. 

4.  If  steamships  are  of  the  same  shape,  their  tonnages  vary  as  the 
cubes  of  their  lengths.     The  Mauretania  is  790  feet  long,  with  a  net 
tonnage  of  32,500.     Construct  a  graph  representing  the  tonnage  of 
ships  of  the  same  shape,  and  of  various  different  lengths. 

Other  ships  which  at  one  time  or  another  have  held  ocean  records 
are  :  the  Deutschland,  length  686  ft.  and  tonnage  16,500  ;  the  Kaiser 
Wilhelm  Der  Grosse,  length  648  ft.  and  tonnage  14,300  ;  the  Lucania, 
length  625  ft.  and  tonnage  13,000  (nearly)  ;  and  the  Etruria,  length 
520  ft.  and  tonnage  8000.  By  means  of  this  graph  decide  whether 
or  not  these  boats  have  greater  or  less  tonnage  than  the  Mauretania 
as  compared  with  their  lengths. 

NOTE.  The  following  more  general  problems  further  illustrate  the 
wide  range  of  application  of  the  fundamental  theorem  that  the  sur- 
faces and  the  volumes  of  similar  solids  vary  respectively  as  the  squares 
and  the  cubes  of  their  corresponding  linear  dimensions.  These  prob- 
lems need  not  be  solved  by  means  of  graphs. 

5.  Raindrops   as  they  start  to  fall  are  extremely  small.     In  the 
course  of  their  descent  a  great  many  are  united  to  form  larger  and 
larger  drops.     If  1000  such  drops  unite  into  one,  what  is  the  ratio  of 
the  surface  of  the  large  drop  to  the  sum  of  the  surfaces  of  the  small 
drops  ? 


VAEIABLE  GEOMETRIC  MAGNITUDES.  167 

6.  Consider  two  machines  similar  in  shape  and  of  heights  hl  and 
A2.     Since  the  tensile  strength  of  corresponding  parts  varies  as  the 
areas  of  their  cross  sections,  it  follows  that  the  tensile  strengths  of 
corresponding  parts  are  in  the  ratio  h^ :  h22.     But  t^ie  total  volumes 
of  the  machines  vary  as  the  cubes  of  the  heights ;  that  is,  the  total 
weights  are  in  the  ratio  h^  :  A23.     Does  this  fact  offer  any  hindrance 
to  the  building  of  machines  indefinitely  large  ? 

7.  The  strength  of  a  muscle  varies  as  its  cross-section  area,  which 
in  turn  varies  as  the  square  of  the  height  or  length  of  an  animal, 
while  the  weight  of  the  animal  varies  as  the  cube  of  its  height  or 
length.     Use  these  facts  to  explain  the  greater  agility  of  small  ani- 
mals.    For  example,  compare  the  rabbit  and  the  elephant. 

8.  Assuming  the  velocities  the  same,  the  amounts  of  water  flow- 
ing through  pipes  vary  directly  as  their  cross-section  areas.     How 
many  pipes,  each  4  in.  in  diameter,  will  carry  as  much  water  as  one 
pipe  72  in.  in  diameter  ? 

9.  What  must  be  the  diameter  of  a  cylindrical  conduit  which 
will  car1>5>  enough  water  to  supply  ten  circular  intakes  each  eight  feet 
in  diameter  ? 

10.  A  water  reservoir,  including  its  feed  pipes,  is  replaced  by 
another,  each  of  whose  linear  dimensions  is  twice  the  corresponding 
dimension  of  the  first.     If  the  velocity  of  the  water  in  the  feed  pipes 
of  the  new  system  is  the  same  as  that  in  the  old,  will  it  take  more  or 
less  time  to  fill  the  new  reservoir  than  it  did  the  old?     What  is  the 
ratio  of  the  new  time  to  the  old? 

11.  If  two  engine  plants  are  exactly  similar  in  shape,  but  each 
linear  dimension  in  one  is  three  times  the  corresponding  dimension 
of  the  other,  and  if  the  steam  in  the  feed  pipes  flows  with  the  same 
velocity  in  both,  compare  the  speeds  of  the  engines. 

12.  If  two  men,  one  5  ft.  6  in.  and  the  other  6  ft.  2  in.  in  height, 
are  similar  in  structure  in  every  respect,  how  much  faster  must  the 
blood  flow  in  the  larger  person  in  order  that  the  body  tissues  of  both 
shall  be  supplied  equally  well  ? 

SUGGESTION.  Note  that  the  amount  of  tissue  to  be  supplied  varies 
as  the  cube  of  the  height,  while  the  cross-section  area  of  the  arteries 
varies  as  the  square  of  the  height. 


CHAPTER  VII. 

• 

THEORY  OF  LIMITS, 

GENERAL  PRINCIPLES. 

311.  In  the  Plane  Geometry  it  was  found  that  there  are 
segments  which  have  no  common  unit  of  measure,  that  is, 
which   are  incommensurable,   and   that   the  ratio   of   the 
lengths  of  two  such  segments  could  be  expressed  only, 
approximately  by  means  of  integers  and   ordinary  frac- 
tions.    Other  incommensurables  occurred  in  dealing  with 
the  length  of  the  circle  and  the  area  inclosed  by  it. 

In  Chapters  III,  IV,  and  V  we  confined  ourselves  to  an 
informal  first  treatment  of  these  incommensurable  ratios, 
tacitly  assuming  their  existence  and  computing  them  ap- 
proximately. In  Chapter  VII  their  existence  was  explicitly 
assumed,  and  certain  theorems  proved  rigorously  on  the 
basis  of  these  assumptions.  In  the  Solid  Geometry  the 
areas  and  volumes  of  the  cylinder,  cone,  and  sphere  have 
been  treated  according  to  this  latter  method. 

312,  In  returning  to  this  subject  once  more  we  fix  our 
attention  on  the  incommensurable  ratios  themselves,  and 
the    method    of   determining   them,   rather   than    on   the 
process  of  approach  and  the  practical  computation  based 
on   it.      We   have  already  used  symbols  such  as    V2   to 
represent  the  ratio  of   the  lengths    of   incommensurable 
segments. 

In  general,  the  ratio  between  any  two  incommensurable 
geometric  magnitudes  of  the  same  kind  may  be  represented 
by  what  is  called  an  irrational  number;  that  is,  a  number 
which  is  neither  an  integer  nor  a  quotient  of  two  integers. 

168 


THEORY  OF  LIMITS.  169 

313,  The  following  method  for  determining  irrational 
numbers  is,  for  simplicity,  applied  first  to  the  integer  1. 

Throughout  this  discussion  the  words  "  point  on  a  line  " 
and  "number"  will  be  used  interchangeably. 

In  a  straight  line  mark  a  certain  point  0  (zero),  and  one  unit  to 

the  right  of  it  mark  another  point  1. 

Also  lay  off  points  such  that  their  0  yz  H  %  l 

distances  from  0  are  |,  f,  |,  £f,  .... 

If  this  sequence  of  points  is  carried  ever  so  far,  it  will  never  reach 
the  point  1.  If,  however,  we  select  a  point  k  to  the  left  of  1,  no  matter 
how  near  it,  we  may  always  go  far  enough  along  this  sequence  to 
reach  points  between  k  and  1. 

314,  The   point   1    has   two   definite   relations   to   this 
sequence. 

(a)  Every  point  of  the  sequence  is  to  the  left  of  1. 

(b)  For  any  fixed  point  K  to  the  left  of  1  there  are 
points  of  the  sequence  between  K  and  1. 

315,  We  now  note  that  1  is  the  only  point  on  the  whole 
line  such  that  both  (a)  and  (5)  are  true  of  it.     For  every 
point  to  the  right  of  1  (a)  is  true,  but  (6)  is  not.     For 
every  point  to  the  left  of  1  (6)  is  true,  but  (a)  is  not. 

It  follows  therefore  that,  while  the  points  of  the 
sequence  merely  approach  the  point  1,  the  sequence,  taken 
as  a  whole,  serves  to  determine  that  point  just  as  definitely 
as  if  the  numeral  1  itself  were  used  to  indicate  the  point. 

316,  The   sequence   3,    2J,    2^,    2|,  ...  is   a   decreasing 
sequence,  and  the  number  2  sustains  relations  to  it  similar 
to  those  described  above.     That  is, 

(a')  Every  point  of  the  sequence  is  to  the  right  of  2. 
(&')  For  every  point  K  to  the  right  of  2  there  are 
points  of  the  sequence  between  K  and  2. 


170  SOLID   GEOMETRY. 

317,  Definitions.     An  endless  sequence  of  the  sort  just 
described  is  called  an  infinite  sequence. 

Not  every  infinite  sequence  serves  to  single  out  a  defi- 
nite point  in  the  manner  shown  above.  Thus  the  sequence 
1,  2,  3,4,  ...  fails  to  do  so,  because  its  terms  grow  large 
beyond  all  bound.  Such  sequences  are  said  to  be  un- 
bounded, while  the  sequence  ^,  f,  |,  ...  is  bounded. 

Again,  the  sequence  1,  2,  1,  2,  1,  ...  fails  to  single  out 
a  definite  point.  This  sequence  is  said  to  be  oscillating, 
since  its  terms  increase,  then  decrease,  then  increase, 
etc.,  while  £,  |,  J-,  ...  is  non-oscillating. 

318,  The  number  1  is  said  to  be  the  least  upper  bound  of 
the  sequence  |,  -J,  -|,  ....     That  is,  1  is  the  smallest  num- 
ber beyond  which  the  sequence  does  not  go.     1  is  also  said 
to  be  the  limit  of  the  sequence. 

Similarly,  2  is  the  greatest  lower  bound  or  the  limit  of 
the  sequence  2|,  2^,  2^,  ...  ;  that  is,  2  is  the  greatest 
number  such  that  the  sequence  contains  no  number  less 
than  it. 

319,  Axiom  X.     Every  bounded  increasing  sequence 
has  a  least  upper  bound,  and  every  bounded  decreasing 
sequence  has  a  greatest  lower  bound. 

This  axiom  may  also  be  stated : 

Every  bounded  increasing  or  decreasing  sequence 
has  a  limit. 

This  axiom  simply  means  that  every  such  sequence 
singles  out  a  definite  number  in  the  manner  stated. 

Thus,  if  we  attempt  to  approximate  the  square  root  of  2,  we  obtain 
a  sequence  1,  1.4,  1.41,  1.414,  1.4142  ... ,  having  for  its  limit  a  definite 
number  represented  by  V2,  which  corresponds  to  the  length  of  the 
diagonal  of  a  square  whose  side  is  unity. 


THEORY  OF  LIMITS.  171 

320,  If  two  sequences  ar  #2,  a3,  ...  and  bv  62,  £>3,  ...  are 
equal  term  by  term,  that  is,  if  ax  =  5j,  #2  =  62,  #3  =  53, . . . 
then  they  are  one  and  the  same  sequence  and  hence  by 
Ax.  X  they  determine  the  same  number. 

The  same  number  may  be  denned  by  two  different 
sequences. 

Thus  each  of  the  sequences,  ^,  f ,  j, ...  and  f ,  f ,  f , ...,  has  1  as  its  limit. 

We  notice,  however,  that  no  matter  what  definite  num- 
ber we  select  in  either  of  these  sequences,  there  is  a  number 
in  the  other  greater  than  it. 

321,  THEOREM.     Two  increasing  bounded  sequences 
define  the  same  number  as  their  limit  if  neither  sequence 
contains  a  number  greater  than  every  number  of  the 
other. 

Let  av  #2,  #3,  ...  and  bv  b2,  53,  ...  denote  two  infinite 
sequences,  with  limits  A  and  J3,  such  that  there  is  no  a 
greater  than  every  5,  and  no  b  greater  than  every  a. 

To  prove  that  A  =  B. 

Proof :  Suppose  that  A  is  not  equal  to  B  and  that  A  is 
less  than  B.  Then  there  must  be  numbers  of  the  sequence 
5j,  62,  63,  •••  greater  than  A,  since  by  §  314  there  are  num- 
bers of  br  ly  b3,"-  greater  than  any  fixed  number  whatever 
which  is  less  than  #,  which  contradicts  the  hypothesis  of 
the  theorem. 

In  like  manner  we  show  that  B  is  not  less  than  A. 
Hence  A  =  B. 

322,  THEOREM.     Two  decreasing  sequences  define  the 
same  number  as  their  limit  if  neither  sequence  contains 
a  number  less  than  every  number  of  the  other. 


172  SOLID   GEOMETRY. 

APPLICATION  OF  LIMITS  TO  GEOMETRY. 

323,  PROBLEM.  On  a  given  segment  AB  to  lay  off  a 
sequence  of  points  #1?  #2?  #3?  of  which  B  is  the  limit, 
such  that  each  of  the  segments  AB{,  AB^  ABS,  •••  is 
commensurable  with  a  given  segment  CD. 

C  mz § D 

ins  wti 


Solution.  Using  mv  an  exact  divisor  of  CD,  as  a  unit 
of  measure,  lay  off  on  AB  a  segment  ABV  such  that  the 
remainder  B^B  is  less  than  mv  Then  CD  and  ABl  are  com- 
mensurable. 

Using  a  unit  w2,  likewise  a  divisor  of  CD,  and  less  than 
.BjB,  lay  off  AB%  such  that  B^B  is  less  than  m2.  Then 
AB2  >  ABV  and  CD  and  AB2  are  commensurable. 

Continuing  in  this  manner,  using  as  units  of  measure 
segments  m3,  m4,  •••  each  an  exact  divisor  of  CD  and  each 
less  than  B%  B,  B±B,  •••  respectively,  we  obtain  a  sequence 
of  segments  ABV  AB2,  ABy  •--,  each  greater  than  the  pre- 
ceding and  each  commensurable  with  CD. 

If  the  units  mv  m2,  m3,  •••  are  so  selected  that  they 
approach  zero  as  a  limit,  it  follows  that  B  is  the  limit  of 
the  sequence  Bv  B^  J53,  •••. 

If  a  different  sequence  of  divisors  of  CD,  as  mi',  m2',  msf  •»,  is  used, 
we  obtain  a  sequence  BI,  B2f,  B3'  ...,  likewise  satisfying  the  conditions 
of  the  problem. 

We  note  in  regard  to  any  two  such  sequences  B\,  B^  Bs,  •••and 
BI',  B%',  BS',  •••  determined  as  above,  that  they  are  both  increasing  and 
that  each  is  such  that  no  point  of  it  is  to  the  right  of  every  point  of  the 
other. 

Hence,  by  §  321  any  two  such  sequences  have  the  same  limit  B. 


. 

THEORY  OF  LIMITS.  173 

324.  If  two  arcs,  AB  and  CD,  of  the  same  circle  or  of 
equal  circles,  are  given,  then,  in  the  same  manner  as  above, 
points  Bv  B2,  B3,  -"  may  be  constructed  on  arc  AB,  forming 
a  sequence  whose  limit  is  B,  such  that  the  arc  CD  is  com- 
mensurable with  every  arc  of  the  sequence  ABV  AB2, 


o  A         o  c 

325,  Definitions.     If  Bv  #2,  B3,  •••  is  a  sequence  of  points 
on  the  segment  AB  having  the  limit  B,  then  the  segment 
AB  is  said  to  be  the  limit  of  the  segments  ABr  AB2,  AB3,  ••-. 

326,  The  ratio  of  two  commensurable  segments  has  been 
defined  as  the  quotient  of  their  numerical  lengths. 

The  ratio  of  two  incommensurable  segments  has  not  been 
explicitly  denned,  but  it  is  now  possible  to  do  so  in  terms 
of  the  limit  of  a  sequence. 

Consider  two  incommensurable  segments  AB  and  CD. 
Let  ar  a2,  ay  •••  be  the  lengths  of  the  segments  each  com- 
mensurable with  CD,  forming  a  sequence  whose  limit  is  the 
segment  AB,  and  let  b  be  the  length  of  the  segment  CD. 

Then  -1,  -2,  -3,  •••  is  an  increasing  bounded  sequence 
666 

having  a  limit  which  we  call  R. 

If  a/,  #2',  a%,  -•  are  the  lengths  of  another  sequence  of 

segments  whose  limit  is  AB,  the  sequence  ^-,  ^2_,  ^-,  ••• 

666 

is  another  increasing  bounded  sequence  with  limit  R'. 

By  §  321  we  now  know  that  R  =  Rf. 

This  number  R  is  defined  as  the  ratio  of  the  segments 
AB  and  CD. 


174 


SOLID  GEOMETRY. 


327,  The  application  of  the  theory  of  limits  to  geometry 
consists  chiefly  in  showing  that  two  numbers  are  equal 
because  they  are  the  limit  of  the  same  sequence,  or  of 
sequences  having  the  property  stated  in  the  theorem  of 
§  321.     In  the  following  paragraphs  the  applications  are 
made  to  the  chief  cases  both  in  plane  and  in  solid  geome- 
try. 

328,  THEOREM.    In  the  same  circle  or  in  equal  circles 
the  ratio  of  two  central  angles  is  the  same  as  the  ratio 
of  their  intercepted  arcs. 


o 


c 


Proof :  In  case  the  arcs  are  commensurable  the  proof 
is  obvious.  See  §  413,  Plane  Geometry. 

If  the  arcs  AB  and  CD  are  not  commensurable,  let  AB^ 
AB2,  AB3,  •-•  be  a  sequence  of  arcs  whose  limit  is  AB,  each 
arc  being  commensurable  with  the  arc  CD. 

Then  the  sequence  — *,  — 2,  — a 

CD       CD       CD 

by  definition,  is  the  ratio  of  the  arcs  AB  and  CD. 


.. has alimitE which, 


Similarly  the  sequence 


Z  AOB,      Z  AOB*     Z  AOB, 


Z  COD  '     Z  COD        Z  COD 

has  a  limit  Ef  which,  by  definition,  is  the  ratio  of  Z  AOB 
and  Z  COD. 

Z-AOB, 


Since 


Z  COD  '     CD        Z  COD  ' 


these   two  sequences  are  identical  and  hence  define  the 
same  limit.     Therefore  it  follows  that  R  =  Rf. 


THEORY  OF  LIMITS. 


175 


329,  THEOREM.  A  line  parallel  to  the  base  of  a  tri- 
angle, and  meeting  the  other  two  sides,  divides  them  in 
the  same  ratio. 


Given  the  A  ABC  with  DE  II  EC  and 
cutting  AB  and  AC. 

AE 
AC 

Proof:    Consider  the  case  when 

AD  and  AB  are  incommensurable. 

Let  DJ,  z>2,  D3  •  •  •  be  a  sequence  on 

AB  whose 


To  prove  that  — 

AS 


limit  is  D.  Through 
these  points  draw  parallels  to  7?c, 
meeting  AC  in  Er  E%,  E%,  •••. 

Then  E  is  the  limit  of  the  sequence  E^  E^  EB,  •  ••. 

For  suppose  it  is  not,  and  that  there  is  a  point  K  on  AE 
such  that  there  is  no  point  of  Ev  E2,  Ez,  •••  between  K  and 
E.  Then  draw  a  line  parallel  to  BC  through  K,  meeting 
AB  in  H. 

But  there  are  points  of  the  sequence  Dv  D^  D3,  between 
H  and  D,  and  hence  points  of  the  sequence  Er  E2,  Ey  •-•  be- 
tween K  and  E,  which  shows  that  E  is  the  limit  of  Ev  Ev 
EB,  ••• 


Now 


AB 


AEl 
AC 


AD  ^ 
AB 


AC 


AD3 
AB 


AES 
AC 


Hence,  the  two  sequences, 


AD,     AD9    AD* 

-  1    —  2   _  _a 


AE,     AE9     AE» 

—  i    —  2    —  3 


.4!?      AB      AB  AC      AC      AC 

are  identical  and  define  the  same  limit  ; 


that  is, 


AD  _AE 
AB~~AC 


17G 


SOLID   GEOMETRY. 


and  cv  c2  cy  ...  are 
is  the  limit  of  the 


330.  Definitions.  If  av  #2,  «3,  •  •  is  a  sequence  with  limit 
a,  then  kav  kav  kay  •••  is  a  sequence  with  limit  ka. 

If  av  «2,  #3,  •••  and  bv  52,  53,  •••  are  two  sequences  with 
limits  a  and  b  respectively,  then  ab  is  the  limit  of  the 
sequence  a^  a252,  a3£>3,  ••-. 

Similarly  if  av  #2,  a3,  •••,  bv  52,  £>3 
sequences  with  limits  a,  6,  c,  then 
sequence  a^b-^c^  &<}><£  y  ^^y  '"• 

For  a  complete  treatment  it  would  be  necessary  to  show  that  these 
definitions  of  multiplication  of  irrational  numbers  are  consistent  with 
the  rest  of  arithmetic  and  also  that  these  new  sequences  are  such  as 
to  determine  definite  limits.  That,  however,  is  beyond  the  scope  of 
this  book. 

If  the  sides  of  a  rectangle  are  incommensurable  with 
the  unit  segment,  we  define  its  area  as  follows  : 

Let  av  «2,  a3,  •••  beu  sequence  of  rational  numbers  whose 
limit  is  the  altitude  a,  and  let  bv  52,  £>3,  •••  be  a  sequence 
whose  limit  is  the  base  b. 

Then  the  area  of  the  rectangle  is  the  limit  of  the  sequence 


But  by  definition  the  limit  of  ajt^  a2b2,  «353,  •••  is  the 
product  ab.     Hence  we  have  the 

331,  THEOREM.       The   area   of  a   rectangle   is    the 
product  of  its  base  and  altitude. 

332,  Definition.     In  a  circle  inscribe  a  sequence  Px,  P2, 
P3,  •••  of  regular  polygons,  each  having  twice  the  num- 
ber of  sides  of  the  one  preceding  it. 


THEORY  OF  LIMITS.  177 

Let  the  perimeters  of  these  polygons  be  pv  p^  p3,  ••• 
and  their  areas  Av  Av  A^  •••.     Then 
the   length   c   of    the  circle  is   the 
limit  of   the  sequence  pr  p%,  py  •  •• 
and  its  area  A  is  the  limit  of  Av  A^ 

A*  •"• 

The  sequence  of  polygons  thus 
inscribed  is  called  an  approximating 
sequence  of  polygons. 

That  these  sequences  are  increasing  and  bounded  is  obvious  from 
the  figure. 

333.  THEOREM.  The  lengths  of  two  circles  are  in  the 
same  ratio  as  their  radii,  and  their  areas  are  in  the  same 
ratio  as  the  squares  of  their  radii. 

Proof  :  Let  the  radii  of  the  circles  whose  centers  are  O 
and  o'  be  r  and  /.  Denote  —  by  k.  Then  rf  =  kr. 

Inscribe  in  O  an  approximating  sequence  of  polygons 
with  perimeters  p^  p^  py  •••  and  areas  A^  A^  A3,  •••.  In 
O!  inscribe  a  sequence  of  similar  polygons.  By  §§  347, 
348,  Plane  Geometry,  the  perimeters  of  the  latter  are  kp^ 
kp2i  kpy  •••  and  their  areas  are  k2Av  k2A2,  k2A%,  •••. 

By  §  330,  if  the  limit  of  pv  pv  ps,  -••  is  c  and  the  limit 
of  A-p  Ay>  Ay  •••  is  A,  then  the  limits  of  kp-^  kp^  kp^  •••, 
and  k2Av  k2A2,  k2A3,  •••  are  kc  and  k2A,  respectively. 

That   is,   the   ratio   of    the   lengths    of    the   circles   is 

^  =  k  =  -,  and  the  ratio  of  their  areas  is  —  =  k2  =  — . 
c  r  A  r2 

If  regular  circumscribed  polygons  are  used,  we  obtain  decreasing 
sequences  both  for  the  perimeters  and  the  areas.  That  the  limits  of 
these  sequences  are  identical  with  those  obtained  above  is  a  direct 
consequence.  See  §  417,  Plane  Geometry, 


178 


SOLID   GEOMETRY. 


334.  Definition.     If  the  three  concurrent  edges  a,  £,  c  of 
a  rectangular  parallelopiped   are   incommensurable  with 
the  unit  segment,    the  volume  in- 
closed is  denned  as  follows  : 

Let  av  dy,  as,  •••  be  a  sequence  of 
rational  numbers  whose  limit  is  the 
side  a.  Let  b^  52, 53,  •  •  •  and  Cj,  <?2,  c3,  •  •  •, 
be  similar  sequences  whose  limits  are 
respectively  the  dimensions  b  and  c?. 

Then  the  volume  is  the  limit  of 
the  sequences  a^c^  ajb^y  as^c^  •••• 

But  the  limit  of  this  sequence  is  by  definition  the 
product  of  the  limits  of  the  three  sequences  «1?  «2,  a3,  •••, 
bv  b2,  63,  •••,  cv  Cy  Cp  ••-,  or  a6c. 

Hence,  we  have  the 

335.  THEOREM.     The  volume  inclosed  l>y  a  rectan- 
gular parallelopiped  is  equal  to  the  product  of  its  three 
concurrent  edges. 

336.  Definition.     In  a  triangular  pyramid  inscribe  and 
circumscribe     prisms 

as  shown  in  the  fig- 
ure. See  §  136  for 
description  of  the 
construction. 

Denote  the  sum  of 
the  volumes  of  the 
inscribed  prisms  by 

n- 

Using  as  altitudes  one  half  the  altitudes  of  the  first  set 
of  prisms,  inscribe  a  second  set  the  sum  of  whose  volumes 
is  F2.  Continuing  in  this  manner,  we  obtain  a  sequence 
of  sets  of  prisms  with  volumes  Vv  Fa,  F3,  •••.  The  limit  F 
of  this  sequence  we  define  as  the  volume  of  the  pyramid. 


THEORY  OF  LIMITS.  179 

If  circumscribed  prisms  are  used,  we  get  a  decreasing  sequence  of 
volumes  F/,  F2',  F3',  •••  with  limit  F'.  That  these  two  limits  are 
identical  follows  from  the  fact  that  the  sum  of  the  volumes  of  the 
circumscribed  prisms  exceeds  that  of  the  inscribed  prisms  by  exactly 
the  volume  of  the  lowest  circumscribed  prism  (see  §  139),  and  this 
may  be  made  as  small  as  we  please. 

337,  EXERCISE. 

1.  Prove  that  the  definition  of  §  334  gives  the  same  volume  for  the 
rectangular  parallelepiped,  no  matter  what  sequences  of  segments  are 
used,  provided  their  limits  are  the  segments  a,  b,  c. 

A  proof  that  the  definition  of  §  336  gives  the  same  volume  for  the 
triangular  pyramid  no  matter  into  how  many  equal  parts  the  altitude 
is  first  divided  is  rather  too  complicated  to  be  attempted  here.  It 
may  be  proved  by  showing  that  if  the  altitude  is  divided  into  n  and 
also  into  m  equal  parts,  then  if  w  >  m,  the  set  of  n  —  1  prisms  will  be 
greater  than  the  set  of  m  -  1  prisms. 

338.  THEOREM.     Two  pyramids  ivith  the  same  alti- 
tudes and  equal  bases  inclose  equal  volumes. 

Proof:  Divide  the  two  equal  altitudes  into  the  same 
number  of  equal  parts  and  construct  inscribed  prisms  as 
in  §  136. 

Since  corresponding  sets  of  prisms  have  equal  volumes, 
it  follows  that  the  volumes  of  the  pyramids  are  the  limits 
of  the  same  sequence,  and  hence  identical. 

339,  Convex   curves.     Without   specifically  defining   a 
convex  closed  curve  (see  §  97),  we  assume  that  in  such  a 
curve  it  is  possible  to  inscribe  a  sequence  of  polygons 
Pv  P2,  P3,  •••,  having  perimeters  pv  jt?2,  p#  •••  and  areas 
Av  A^  Ay  •••  with  limits  p  and  A  respectively,  and  to  cir- 
cumscribe a  sequence  of  polygons  P/,  P2',  P3',  •••,  having 
perimeters  p^,  p^p^  -•  and  areas  A^,  A2',  A3',  •••  with 
limits  p'  and  Af  respectively,  such  that  p  —  p'  and  A  =  A1. 

These  limits  p  and  A  we  now  define  as  the  perimeter  and 
the  area  respectively  of  the  curve. 


180  SOLID   GEOMETRY. 

340,  Definition.      Given   any  cylinder  with   a   convex 
right  cross  section  and  an  element  e.     In  this  cross  section 
inscribe  a  sequence  Pv  P2,  P3,  •••  of  polygons,  as  in  §  332, 
with  perimeters  pv  pv  p^  •  ••  and  areas  Av  ,42,  J3,  ••-,  thus 
denning  the  perimeter  p   and  the  area   A  of   the  cross 
section. 

Consider  a  set  of  prisms  inscribed  in  this  cylinder,  of 
which  PJ,  P2,  Pg,  ...  are  right  cross  sections. 

Then  the  areas  and  the  volumes  of  these  prisms  are 
respectively  p^,  p2e,  p5e,  •••  and  A^,  A2e,  A^e,  •». 

The  lateral  area  and  the  volume  of  the  cylinder  are  now 
defined  as  the  limits  of  these  sequences.  But  these  limits 
are  by  §  330  equal  to  pe  and  Ae  respectively. 

Hence,  we  have  the 

341,  THEOREM.     The  lateral  area  of  a  cylinder  is 
the  product  of  an  element,  and  the  perimeter  of  a  right 
section  and  its  volume  is  the  product  of  an  element  and 
the  area  of  a  right  section. 

342,  EXERCISES. 

1.  Give  examples  other  than  those  given  in  the  text  of  infinite 
sequences  which  do  not  determine  definite  numbers. 

2.  Give  two   distinct  sequences  which  determine  the  number  2. 
Show  that  the  theorem  of  §  321  applies  and  proves  that  these  se- 
quences determine  the  same  number. 

3.  Give  two  decreasing  sequences  each  of  which  determines  the 
number  3.     Apply  §  322  to  show  that  these  sequences  determine  the 
same  number. 

4.  State  fully  the  relation  between  a  bounded  increasing  sequence 
and  the  number  determined  by  it.     State  also  the  relations  between 
a  bounded  decreasing  sequence  and  the  number  determined  by  it. 

5.  State  fully  what  is  meant  by  "  a  limit  of  a  sequence  "  both  for 
increasing  and  decreasing  sequences. 


THEORY  OF  LIMITS.  181 

6.  Given  two  incommensurable  segments  AB  and  CD.   Layoff  on 
the  line  AB  a  decreasing  sequence  of  segments,  each  of  which  is  com- 
mensurable with  CD,  such  that  the  limit  of  the  sequence  is  the  seg- 
ment AB. 

7.  If  aij  a-z,  as,  •••  is  an  increasing  sequence  denning  the  number 
4,  prove  that  3  ai,  3  «2,  3  «s,  •••  defines  the  number  3  x  4  =  12. 

Note  that  in  case  the  sequence  ai,  a2,  as,  •  •  •  defines  an  irrational 
number,  we  should  not  be  able  to  prove  the  corresponding  proposition 
without  first  defining  what  is  meant  by  the  product  of  a  rational  and 
an  irrational  number.  But  such  definition  would  in  that  case  be  the 
proposition  itself.  See  §  330. 

8.  If  ai,  02,  ag,  •••  and  &1?  &2>  bs,  •••  are  increasing  sequences  defin- 
ing the  numbers  3  and  5,  show  that  the  sequence  a\b\,  a^b-z,  a^bs,  •••  de- 
fines the  number  15. 

In  case  these  sequences  defined  irrational  numbers,  could  a  cor- 
responding proposition  be  proved? 

9.  In  the  same  manner  as  in  §  326  define  the  ratio  of  two  incom- 
mensurable arcs. 

10.  Show  as  above  that  the  ratio  so  obtained  is  independent  of  the 
sequence  of  units  of  measurement  used,  so  long  as  the  limit  of  this 
sequence  is  zero. 

11.  Treat  the  ratio  of  two  incommensurable  angles  in  a  manner 
similar  to  the  treatment  of  arcs  in  the  two  preceding  exercises. 

12.  Define  the  lateral  area  of  a  cylinder  by  means  of  circumscribed 
prisms,  and  show  that  this  definition  leads  to  the  same  result  as  that 
given  in  §  340. 

13.  Prove  as  above  that  the  volume  of  any  convex  cylinder  is  equal 
to  the  product  of  its  altitude  and  area  of  its  base. 

14.  Prove  that  the  lateral  area  of  a  right  circular  cone  is  equal  to 
half  the  product  of  the  slant  height  and  the  perimeter  of  its  base. 

15.  Prove  that  the  volume  of  any  convex  cone  is  equal  to  one  third 
the  product  of  its  altitude  and  the  area  of  its  base. 

SUGGESTION.  The  treatment  required  in  the  last  three  exercises  is 
a  very  close  paraphrase  of  the  definitions  and  proof  given  in  §  334. 
Observe  that  we  cannot  begin  to  make  a  proof  until  we  have  defined 
the  subject  matter  of  the  theorem.  That  is,  we  must  first  define  the 
areas  and  volumes  in  question. 


182 


SOLID   GEOMETRY. 


VOLUME  OP  THE  SPHERE. 

343,    Through  two  points  P  and  Q  of  a  sphere  pass  a 
great  circle  forming  a  hemisphere  with  center  o. 


Divide  OA,  the  radius  perpendicular  to  the  plane  of 
the  great  circle,  into  the  equal  parts  O<7,  CB,  and  BA. 
Through  C  and  B  pass  planes  parallel  to  the  plane  of 
POQ,  meeting  the  circle  in  points  D  and  E,  respectively. 

Construct  right  circular  cylinders  with  axes  OC  and  CB 
and  radii  CD  and  BE.  Denote  by  vl  the  sum  of  the 
volumes  of  these  cylinders. 

Now  divide  the  radius  OA  into  six  equal  parts  and  con- 
struct five  cylinders  in  the  same  manner  as  above.  Let 
the  sum  of  these  volumes  be  F2. 

Continuing  in  this  manner,  each  time  dividing  OA  into 
twice  as  many  equal  parts  as  in  the  preceding,  we 
obtain  a  sequence  of  sets  of  cylinders  and  a  corresponding 
sequence  Fx,  F2,  F3,  •••  of  volume. 

We  now  define  the  volume  inclosed  by  the  hemisphere  as 
the  limit  of  the  sequence  Fr  F2,  F3,  •••. 

344,  Construct  a  right  circular  cylinder  with  its  base  in 
the  plane  of  POQ  and  altitude  o' A'  =  OA. 

Denote  by  F  the  figure  formed  by  the  lower  base  of  the 
cylinder,  its  lateral  surface  and  the  lateral  surface  of  the 
cone  whose  base  is  the  upper  base  of  the  cylinder  and 
whose  vertex  is  at  o' . 


THEORY  OF  LIMITS.  183 

Draw  segments  O'M  and  O'N.  Let  the  planes  through 
C  and  B  cut  OrAf  in  c1  and  B'  and  o'Jlf  in  H  and  .HT. 

Now  form  the  cylinder  O'C'H  whose  axis  is  OrCf  and 
whose  radius  is  C'H.  Likewise  form  C'B'K. 

Let  F/  denote  the  sum  of  the  volumes  of  o'cfDf  and 
C'B'E'  minus  the  sum  of  the  volumes  O'C'H  and  C'B'K. 

In  a  similar  manner,  using  the  planes  which  divide  OA 
and  hence  o'Ar  into  six  equal  parts,  we  form  another  set  of 
five  cylinders,  the  sum  of  whose  volumes  minus  that  of  the 
smaller  inside  cylinders  we  denote  by  F2'. 

Continuing  in  this  manner,  we  obtain  a  sequence  of  vol- 
umes F/,  F2',  F3',  •••  whose  limit  v'  we  define  as  the  volume 
of  the  given  figure  F. 

We  now  prove  that  Vl  =  F/,  F2  =  F2',  •••. 
Denote  OA  by  r,  and  note  that  OfBr  =  B'K. 

(1)  Vol.^f&E'  -  Vol.  C'B'K  =  TTC'B'  (B7!?'2 

7rC'B'(r2-o'B'2). 

(2)  Vol.  CBE  =  TTCB  •  BE2  =  TTCB  (r2  —  OB2),  since  BE2  = 
r2  —  OB2. 

But  OB  =  O'B'  and  CB  =  C'B'. 

Hence,  Vol.  CBE  =  Vol.  C'B'E'  -Vol.  c'^'^:. 

Similarly  we  show  that 

Vol.  OCD  =  Vol.  O'C'D'  —Vol.  O'C'H. 
Hence,  F:  =  F/.     In  like  manner  F2  =  F2',  F3=  F3',  •••• 
Hence,  F  =  F',  since  they  are  the  limits  of  the  same 
sequences. 

But  the  volume  of  the  cylinder  O'A'M  is  Trr3  and  of  the 

cone  whose  volume  was  subtracted,  —  r3-    That  is,  the  vol- 

2  TT  2  TT 

ume  of  F  is——  r3,  and  hence  that  of  the  hemisphere  is  -—  r3. 
o  o 

Hence,  we  have  the 

345,    THEOREM.     The  volume  of  the  sphere  is  -|  Trr3. 


184  SOLID   GEOMETRY. 

346,  Note  that  the  above  proof  consists  essentially  in 
showing  that  the  area  of  the  circle  BE  is  equal  to  that  of 
the  ring  between  the  circles  BrEf  and  B1 K,  and  that  the 
area  of  the  circle  CD  is  equal  to  that  of  the  ring  between 
C'D'  and  C'H  and  so  on. 

Indeed  this  theorem  and  also  that  of  §  139  are  special 
cases  of  what  is  known  as 

Cavalieri's  Theorem.  If  two  solid  figures  are  re- 
garded as  resting  on  the  same  plane  b,  and  if  in  every 
plane  parallel  to  1}  the  sections  of  the  two  figures  have 
equal  areas,  the  figures  have  equal  volumes. 

The  proof  of  this  more  general  theorem  is  more  difficult  than  any 
thus  far  given,  inasmuch  as  it  involves  sequences  which  oscillate  ;  that 
is,  which  are  neither  constantly  increasing  nor  constantly  decreasing. 

THE  AREA  OF   THE   SPHERE. 

347,  About  a  sphere  of  radius  r  construct  a  sequence  of 
circumscribed  polyhedrons  such  that  the  largest  face  in 
each  polyhedron  becomes  as  small  as  we  please  when  we 
proceed  along  the  sequence.     Let  sv  «2,  s3,  •••  be  the  total 
surfaces  of  these  polyhedrons.     This  forms  a  decreasing 
sequence  with  limit  8  which  we  define  as  the  surface  of  the 
sphere. 

The  volumes  of  these  polyhedrons  will  be  -sv  -s2,  -s3,  ••-. 

ooo 

Then  the  volume  V  of  the  sphere  is  denned  as  the  limit 
of  this  sequence  of  volumes. 

Hence,  by  §  330,    v  =  f  8.     But  by  §  245,  V  =  |  Trr3. 

o 

Then  8  =  --  47rr3=47rr2. 

r     3 

Hence,  we  have  the 

348,  THEOREM.      The  area  of  the  sphere  is  4  Trr2. 


THEOEY  OF  LIMITS  185 

This  argument  is  incomplete  in  that  two  distinct  defi- 
nitions have  been  given  of  the  volume  of  the  sphere, 
namely,  as  the  limit  of  an  increasing  sequence  of  volumes 
of  inscribed  cylinders  and  also  as  the  limit  of  a  decreasing 
sequence  of  volume  of  circumscribed  polyhedrons.  But 
it  has  not  been  proved  that  these  two  definitions  are  equiva- 
lent ;  that  is,  that  they  lead  to  the  same  formula  for  the 
volume  of  the  sphere. 

The  treatment  of  the  area  and  volume  of  the  sphere 
in  most  of  the  current  text-books  on  geometry  is  open 
to  a  similar  objection.  Usually  two  distinct  definitions 
are  used  for  the  area  of  the  sphere,  namely,  as  a  limit  o£ 
the  areas  of  circumscribed  frustums  of  cones  and  also  as  the 
limit  of  the  surfaces  of  circumscribed  polyhedrons. 

The  argument  used  in  §§  285-293  is  not  subject  to  any 
such  objection. 

SUMMARY  OF  CHAPTER  VII. 

1.  Explain  the  separate  stages  of  treatment  of  incommensurables 
and  of  limits  as  given  in  this  text,  including  both  the  Plane  and  Solid 
Geometry. 

2.  Give  an  outline  of  the  introduction  to  this  final  treatment  by 
means  of  sequences. 

3.  Make  a  list  of  the  definitions,  principles,  and  theorems  upon 
which  the  treatment  of  sequences  is  based. 

4.  Explain  fully  the  way  in  which  the  principles  of  sequences  are 
applied  to  geometric  theorems.     Illustrate  by  line-segments  and  arcs. 

5.  Make  a  list  of  the  theorems  which  are  proved  here  by  the 
theory  of  limits,  using  the  sequence  process. 

6.  State  in  some  detail  how  this  process  is  used  in  case  of  the  area 
and  volume  of  the  sphere! 

7.  Give  all  the  mensuration  formulas  proved  in  this  chapter. 

8.  Review  the  entire  list  of  mensuration  formulas  for  both  Plane 
and  Solid  Geometry. 


INDEX. 


(References  are  to  sections  unless  otherwise  stated.) 


Altitude,  of  a  cone 149 

of  a  frustum 144 

of  a  prism 71 

of  a  pyramid 131 

of  a  spherical  segment     .     .     .  295 

of  a  zone 295 

Angle,  between  two  curves   .     .  223 

dihedral 40 

face 63 

of  projection 116,125 

polyhedral 63 

spherical 223 

trihedral 63 

Approach 312 

Arc  of  a  spherical  triangle     .     .  225 

Area,  of  a  cone 157 

of  a  cylinder  ....   107,  108,  340 

of  a  prism 73 

of  a  pyramid 132 

of  a  rectangle 330 

of  a  sphere      .  285,  290,  291,  347,  348 

of  a  spherical  polygon      .     .     .  283 

of  a  spherical  triangle     .     .     .  281 

of  a  zone 296 

lateral 71 

total 71 

Axioms,  5,  7,  9,  85,  105,  107,  137, 

157,  290,  319 

Axis,  in  a  graph    .     .     .     Chapter  VI 

of  a  circle 204 

Base,  of  a  cone 149 

of  a  cylinder 99 

of  a  prism 70 

of  a  pyramid 131 

of  a  spherical  sector    ....  295 

of  a  spherical  segment     .     .     .  295 

Bir octangular  spherical  tri- 
angle        278 


Bisector  of  dihedral  angle    .  54 

Bound,  greatest  lower  ....  318 

least  upper 318 

Cavalieri's  theorem  ....  346 

Center,  of  a  sphere 200 

of  similitude -  .  188 

Circle,  axis  of 204 

great 204 

poles  of 204 

small 204 

Circular,  cone 149 

cylinder 99 

Circumscribed,  cone  ....  155 

cylinder 106 

polyhedron 182 

prism 100,  130 

pyramid 155 

sphere 177,  217 

Commensurable,  angles,  arcs  328 

ratio 326 

segments .  326 

Cone,  altitude  of 149 

base  of 149 

circular 149 

element  of 148, 151 

lateral  surface  of  ....  149, 162 

oblique 149 

right  circular 149 

vertex  of 148 

volume  inclosed  by  ....  167 

Congruent,  polyhedral  angles  66 

polyhedrons 89 

spherical  polygons 276 

Conical  surface,  directrix  of  .  148 

element  of ........  148 

generator  of 148 

nappes  of 148 

vertex  of  .  ,  148 


187 


188 


INDEX. 


(References  are  to  sections  unless  otherwise  stated.) 


Corresponding,   cross-sections  199 

linear  dimensions 199 

parts  of  polar  triangles   .     .     .  252 

parts  of  similar  polyhedrons    .  183 

points  in  similitude      ....  188 

Cosine  of  an  angle      ....  117 

Cube 79 

Curved  surface 97,  104 

Curves,  angle  between      ...  223 

convex 97 

Cylinder,  bases  of 99 

circular 99 

circumscribed 106 

inscribed 106 

lateral  surface  of    ....  99,  107 

of  revolution 99 

radius  of 99 

right 99 

volume  of 107, 112 

Cylindrical  surface,  directrix 

of 98 

element  of 98 

generator  of 98 

Degree,  spherical 278 

Dihedral  angles 40 

adjacent 45 

bisector  of 54 

complementary 45 

equal 42 

measure  of 45 

supplementary 45 

vertical 45 

Distance,   between   two  points 

on  a  sphere 206 

from  a  point  to  a  plane   ...  20 

polar 208 

ratio 190 

Dodecagon 170,181 

Edge,  lateral 70 

of  a  dihedral  angle 4C 

of  a  polyhedral  angle  ....  62 

of  a  polyhedron 

Element,  of  a  cone 148 

of  a  cylinder 9i 

Ellipse,  area  of 

definition  of 127 

Equal  solids 89 


Equivalent  solids 
Euler's  theorem 


172 


Faces,  of  a  dihedral  angle     .     .  40 

of  a  polyhedral  angle  ....  63 

of  a  polyhedron 68 

of  a  prism 70 

of  a  pyramid 131 

Figure  in  space 3 

Frustum 144 

Generator,  of  a  conical  surface  148 

of  a  cylindrical  surface   ...  98 

of  a  pyramidal  surface    .     .     .  130 

of  a  spherical  surface  ....  289 

Geometrical  solid 2 

Graphic  representation, 

Chapter  VI 

Great  circle,  axis  of    ....  204 

pole  of 208 

of  a  sphere 204 


Half-plane 


40 


Icosahedron 170, 181 

Incommensurables    .     86, 311, 312 

Inscribed,  cone 155 

cylinder 106 

prism 106,137 

pyramid 155 

sphere    ....     177,178,179,217 

Irrational,  number 312 

ratio 312 

Isosceles  spherical  triangle    239 

Lateral,  edges 70 

faces 70 

surfaces 131 

Limit,  of  segments 325 

of  a  sequence 318, 319 

Loci  problems,  pages  5,  10,  11, 

15,  16,  17,  20,  29,  30,  31,  118,  123 

Lune,  angle  of 278 

Measurement,  of  surfaces,  104, 

155,  285 
of  volumes      82, 89, 104, 138,  155,  285 


INDEX. 


189 


(References  are  to  sections  unless  otherwise  stated.) 


Nappes,  of  a  conical  surface      .    148 
of  a  pyramidal  surface    .     .     .    130 

Octahedron 170, 181 

Pantograph 198 

Parallel,  line  to  a  plane     ...      29 
plane  to  a  plane 29 

Parallelepiped,  rectangular     .      79 
volume  inclosed  by       ...      82-86 

Perpendicular,  line  to  a  plane      13 

plane  to  a  line 13 

plane  to  a  plane 47 

Plane,  determination  of     ...        8 

projections  upon 125 

-segment 67 

Plane  angle  of  a  dihedral  angle      40 

Polar,  distances 208 

triangle 249 

Poles  of  a  circle 204 

Polygonal  plane-segments    .     .      67 

Polygons,spherical,congruent    276 
symmetrical 232 

Polyhedral  angles  ....   63,  271 

congruent 65 

edges  of 63 

faces  of 63 

symmetrical 232 

Polyhedrons,  added    ....      75 

circumscribed 217 

congruent 89 

corresponding  parts  of     ...    183 

edges  of 68 

equal,  equivalent 

faces  of  

inscribed 176, 217 

regular 169, 170 

similar 183 

vertices  of 68 

Prismatic  surface,  directrix  of      69 
generator  of 69 

Prisms,  altitude  of 71 

area  of 71 

bases  of 70 

circumscribed 106, 137 

hexagonal 71 

inscribed 106, 137 

lateral  edges  of 7C 

lateral  faces  of 70 


oblique 71 

quadrangular . 71 

regular 71 

right 71 

triangular 71 

volume  of 90 

Problems  and  Applications 

Pages  28,  53, 66, 79,  91, 112, 156, 166 

Projection,  of  a  circle  ....  127 

of  a  figure 57 

of  a  point 57 

of  a  line-segment 115 

of  a  plane-segment 125 

Projection  angle 116 

Pyramid,  altitude  of    ....  131 

base  of 131 

frustum  of 144 

lateral  faces  of 131 

regular 131 

right 131 

triangular 131 

truncated 144 

Pyramidal  surface     ....  130 

directrix  of 130 

element  of 130 

nappes  of 130 

vertex  of 130 


Quadrant 


208 


Radius,  of  a  circular  cylinder  .  99 

of  a  sphere 200 

Ratio,  commensurable  ....  326 

incommensurable 326 

of  similitude 190 

Regular,  polyhedrons  .  169,  170,  171 

prisms 71 

pyramids 131 

tetrahedrons 170 

Right,  cone 149 

cylinder 99 

prism 71 

pyramid 131 

section 70,  99,  149 

Section,  of  a  prismatic  surface  .  69 

of  a  sphere 203 

Sector,  spherical 295 


190 


INDEX. 


(Keferences  are  to  sections  unless  otherwise  stated.) 


Segment,  altitude  of  ....  295 

bases  of 295 

spherical 295 

Sequence,  approximating  .  .  .  331 

bounded 317 

decreasing 316 

increasing 313 

infinite 317 

limit  of 317 

non-oscillating 317 

oscillating 317 

unbounded 317 

Similar,  cone  of  revolution  .  .  160 
cylinder  of  revolution  ....  110 

figures 189 

polyhedrons 183 

Similarity,  applications  of     .    .  198 

Similitude,  center  of  .  .  188-194 
ratio  of 180 

Slant  height,  of  frustum  .  .  .149 

of  pyramid 131 

of  right  cone 149 

Small  circle  on  a  sphere    .     .     .  204 

Solids,  congruent 65,  276 

equal 89 

equivalent 89 

geometrical 3 

Sphere,  area  of  ...  289-291,  347 

center  of 177, 200 

circumscribed 177,217 

determination  of 179 

diameter  of 200 

great  circle  of 204 

inscribed 182,217 

points  within,  without  .  .  .  200 

radius  of 177, 200 

small  circle  of 204 

tangent  to 214 

volume  of 285-293 

Spherical,  angle 223 

blackboard 211 

degree,  minute,  second  .  .  .  278 

excess 280, 282 

polygons 272,  276 

sector 295,  297 

segment 295,296 

surface 289 

triangle 225 


Summaries,  pages  28, 65,  90, 112, 

155,165,185 

Surface,  conical 148 

curved 97,  104 

cylindrical 98 

lateral 99 

prismatic 69 

pyramidal    ........  130 

spherical 200 

Symmetrical,  spherical  triangles  232 
trihedral  angles 232 

Symmetry  with    respect    to    a 
point 188 

Tangent,  of  an  acute  angle  .  .  117 

to  a  cone 151 

to  a  cylinder 101 

to  a  sphere 214 

Tetrahedron,  circumscribed  .  .  182 

inscribed 131, 176, 177 

regular 170 

Theory  of  limits  .     .      Chapter  VII 

Triangles,  bi-rectangular  .  .  .278 

congruent 276 

isosceles 239 

polar 249 

spherical 225 

symmetrical 232 

Truncated,  prism  .  .  .  .  75 
pyramid 144 

Variables Chapter  VI 

Vertex,  of  a  cone 148 

of  a  polyhedral  angle    ....    63 

of  a  pyramid 130 

of  a  spherical  angle 223 

Vertices,  of  a  polyhedron  ...    68 

Volume,  as  a  limit  of  a  sequence  330 

of  a  cylinder    ....  107, 112,  339 

of  a  prism 86,  91 

of  a  sphere  ....      285,  292,  343 

of  spherical  cone 297 

of  spherical  sector 297 

of  spherical  segment     ....  299 

Zone,  altitude  of 295 

bases  of 295 

of  one  base 295 


14  DAY  USE 


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